🧠:[<think>Okay, so I have this problem here: Given a positive integer n, determine the maximum number of lattice points in the plane a square of side length n + 1/(2n+1) may cover. Hmm, interesting. Let me think through this step by step.First, I need to recall what a lattice point is. A lattice point in the plane is a point with integer coordinates, right? So, for example, (1,2) or (3,4) are lattice points. The question is about a square with a specific side length and how many such points it can cover. The goal is to find the maximum number possible.The side length of the square is given as n + 1/(2n+1), where n is a positive integer. So, for n=1, the side length would be 1 + 1/3 = 4/3, for n=2, it would be 2 + 1/5 = 11/5, and so on. The challenge is to position this square in such a way that it covers as many lattice points as possible.Let me start by considering how squares can cover lattice points. If a square is aligned with the grid (i.e., its sides are parallel to the x-axis and y-axis), then the number of lattice points it covers can be related to its side length. However, intuitively, rotating the square might allow it to cover more lattice points, even if the side length is fixed. So, perhaps the maximum number of lattice points is achieved when the square is rotated at a 45-degree angle? But I need to verify this.Wait, but the side length here is n + 1/(2n+1). For n being a positive integer, the main term is n, and then there's a small fraction added. So the square is just slightly longer than n units. If we have a square of side length n aligned with the grid, it would cover (n+1)^2 lattice points, right? Because each unit square has (n+1) points along each side. But if we rotate the square, the number might be different.But in this case, the square is slightly larger than n. So perhaps when rotated, it can cover more points? Hmm. Let me think.Alternatively, maybe aligning the square with the grid is better. Let's try to analyze both cases.First, consider an axis-aligned square. If the square has side length s, then the number of lattice points it contains is roughly (floor(s) +1)^2. But since the side length here is n + 1/(2n+1), which is just a bit more than n. So, if the square is axis-aligned, starting at (0,0), it would go from (0,0) to (n + 1/(2n+1), n + 1/(2n+1)). But the number of lattice points inside would be the number of integer points (x,y) where 0 ≤ x ≤ n + 1/(2n+1) and 0 ≤ y ≤ n + 1/(2n+1). Since x and y must be integers, the maximum x and y can be is n, because n + 1/(2n+1) is less than n + 1. Therefore, the number of lattice points would be (n+1)^2.But wait, if the square is placed such that its sides are not aligned with the grid, maybe we can have more points. Let me recall that a rotated square can sometimes cover more lattice points. For example, a square rotated at 45 degrees with side length sqrt(2) can cover 5 lattice points: the four corners and the center. But if it's axis-aligned, a square of side length 1 would cover 4 lattice points. So rotation can help.But in our case, the side length is n + 1/(2n+1). Let me compute the diagonal of such a square. The diagonal length would be s*sqrt(2), where s is the side length. So, for s = n + 1/(2n+1), the diagonal is (n + 1/(2n+1)) * sqrt(2). If we rotate the square so that its diagonal is aligned with the axes, maybe the projection along each axis is longer, allowing it to cover more lattice points? Not sure.Alternatively, perhaps the number of lattice points covered by a square is related to its area? But the area here would be s^2 = (n + 1/(2n+1))^2. But according to the Gauss circle problem, the number of lattice points inside a circle of radius r is roughly πr^2 + E, where E is the error term. But for squares, the relationship might be different.However, we are dealing with squares, not circles. The exact number of lattice points in a square can vary depending on its placement. So, perhaps the maximum number is related to the area, but we need a more precise approach.Another idea: For a square of side length n, aligned with the grid, it covers (n+1)^2 lattice points. If we make the side length slightly larger, say n + ε, where ε is a small positive number, then depending on how we position the square, we might be able to include one more row or column of lattice points. However, since ε here is 1/(2n+1), which is a specific value, maybe this allows the square to just barely cover an extra column or row when shifted appropriately.Wait, but if the square is axis-aligned and has side length n + 1/(2n+1), then shifting it by less than 1 unit could potentially include an extra lattice point. But if it's not shifted, then the number of lattice points would still be (n+1)^2 because the upper bound is n + ε < n + 1. So, for example, if you have a square from 0 to n + ε in both x and y, the maximum integer coordinate in each direction is n, so the number of lattice points is (n+1)^2. However, if we shift the square so that it starts near the middle of a lattice point, maybe we can include another column or row?Wait, let's think of an example. Let n=1. Then the side length is 1 + 1/3 = 4/3 ≈ 1.333. If we place the square axis-aligned from 0 to 4/3 in both x and y, then the lattice points inside would be (0,0), (0,1), (1,0), (1,1). That's 4 points. But if we shift the square to, say, start at -1/3 in both x and y, then it would go from -1/3 to 1 in both directions. Then, lattice points would include (-1,-1), (-1,0), (-1,1), (0,-1), (0,0), (0,1), (1,-1), (1,0), (1,1). But wait, the square is only 4/3 in length. If shifted by -1/3, it would cover from -1/3 to 1, which is 4/3 length. But (-1,-1) is at (-1,-1), which is outside the square since the square starts at -1/3. Wait, maybe this shifting idea isn't correct.Alternatively, perhaps placing the square such that it's centered at a lattice point. For example, centering the square at (0.5, 0.5) with side length 4/3. Then the square would extend from 0.5 - 2/3 = -1/6 to 0.5 + 2/3 = 7/6 in both x and y. So, in this case, the square would cover x from -1/6 to 7/6, which is approximately 1.166 in length. Wait, no, 7/6 - (-1/6) = 8/6 = 4/3, which is correct. Now, in this case, the lattice points covered would be (0,0), (0,1), (1,0), (1,1). Still 4 points. Because between -1/6 and 7/6, the integer x-coordinates are 0 and 1. So same as before. So shifting the square's center doesn't help in this case.But maybe if we rotate the square? For example, rotating a square of side length 4/3 by 45 degrees. How would that affect the number of lattice points?A square rotated by 45 degrees with side length s has its vertices at a distance of s*sqrt(2)/2 from the center. So, for s = 4/3, the distance from center to each vertex is (4/3)*(sqrt(2)/2) = (2*sqrt(2))/3 ≈ 0.9428. So, if we center this rotated square at (0.5, 0.5), the vertices would be approximately 0.5 ± 0.9428 in both x and y directions. Wait, but that would make the square extend beyond 1.44 in either direction. So, in terms of lattice points, the x-coordinates covered would be from approximately 0.5 - 0.9428 ≈ -0.4428 to 0.5 + 0.9428 ≈ 1.4428. So the integer x-coordinates covered would be 0 and 1. Similarly for y. So the lattice points would still be (0,0), (0,1), (1,0), (1,1), same as the axis-aligned case. So maybe rotating doesn't help here either.Hmm. So for n=1, the maximum number of lattice points seems to be 4, which is (1+1)^2. But wait, is there a way to position the square to cover more?Wait, maybe if the square is not centered at a half-integer coordinate. Suppose we position the square such that its sides are between lattice points. For example, if the square is placed such that it starts just after a lattice point, maybe it can cover an extra point.Wait, for example, let's take the square from ε to n + ε + 1/(2n+1), where ε is a very small positive number. Then, if ε is small enough, the square would include the lattice points from 0 to n in each dimension, but not n+1. So again, (n+1)^2 points.Alternatively, if we shift the square such that it straddles a lattice line. Wait, maybe overlapping multiple lattice points. Wait, I need to think differently.Wait, another approach: the maximum number of lattice points a square can cover is related to its perimeter. There's a theorem called the theorem of Beck, but I'm not sure. Alternatively, the number of lattice points a convex set can cover is roughly proportional to its area plus its perimeter. But I need to recall the exact result.The general upper bound for the number of lattice points inside a convex set in the plane is given by Blichfeldt's theorem, which states that the number of lattice points is less than the area plus something. But maybe more straightforwardly, for a square, the number of lattice points it can cover is at most (floor(s) + 1)^2, where s is the side length. But if the square is rotated, maybe this formula doesn't apply.Alternatively, consider that each side of the square can intersect lattice lines (vertical or horizontal lines at integer coordinates). The number of intersections relates to the number of lattice points on the boundary. But perhaps that's more complex.Wait, maybe the key here is that the square is just barely longer than n. So, if we can position the square such that it can cover an additional column or row beyond n+1 in each direction. But since the side length is n + 1/(2n+1), which is less than n + 1, perhaps we can't cover an entire extra column. But maybe if the square is placed diagonally, it can cover more lattice points by crossing through multiple rows and columns?Wait, for example, a square of side length n + ε, when rotated, its projections onto the x and y axes are longer. The projection length is s*cosθ + s*sinθ, where θ is the angle of rotation. To maximize the number of lattice points, we might need the projections to cover as many integer intervals as possible.For a square rotated by θ, the width in the x-direction is s*(cosθ + sinθ). Similarly for the y-direction. So, to maximize the number of lattice points, we need this width to be as large as possible. However, the actual number of lattice points would be roughly the product of the number of integer intervals covered in x and y directions. So, if the projection in x is m units and in y is m units, then the number of lattice points would be approximately (m+1)^2.But since the side length is s = n + 1/(2n+1), and we can rotate the square to maximize the projections. The maximum projection occurs at θ=45 degrees, where the projection becomes s*sqrt(2). Wait, but when θ=45 degrees, the projection length is s*(cos45 + sin45) = s*sqrt(2). So for our case, s*sqrt(2) = (n + 1/(2n+1)) * sqrt(2). Let's compute this.For example, take n=1: s=4/3, so projection length is 4/3 * sqrt(2) ≈ 1.885. So the projection is less than 2, so the number of integer intervals covered is 1 in each direction, leading to (1+1)^2=4 lattice points. Which is same as axis-aligned.For n=2: s=2 + 1/5 = 11/5 = 2.2. Projection length is 2.2*sqrt(2) ≈ 3.11. So the projection is just over 3, which would mean that if the square is positioned correctly, the projection could cover 3 units in x and y directions, leading to (3+1)^2=16 lattice points. But the original axis-aligned square would cover (2+1)^2=9 lattice points. Wait, that seems like a big difference. Hmm, but is that correct?Wait, wait. If the square is rotated by 45 degrees and the projection length is ~3.11, does that mean that along the x-axis, the square spans approximately 3.11 units? So, depending on where you place the square, it could cover 3 or 4 integer intervals. But to cover 4 intervals, the projection needs to be at least 3 units. Since 3.11 is just over 3, maybe it can cover 4 intervals if positioned correctly. For example, if the projection starts at 0.5, spanning from 0.5 - 1.555 to 0.5 + 1.555, which is from approximately -1.055 to 2.055. So, the integer points covered would be -1, 0, 1, 2. But if the square is shifted, maybe it can cover 0,1,2,3. Wait, but the actual coordinates depend on the exact placement.But the problem is that the square itself has area (11/5)^2 ≈ 4.84, but the number of lattice points (3+1)^2=16 would be way higher than the area. That seems impossible, since each lattice point corresponds to a unit area. So there must be a mistake here.Wait, I think the projection idea might be flawed. Because when you rotate the square, even though the projection is longer, the actual number of lattice points covered doesn't just depend on the projection. It's more about how the square intersects the grid lines.Alternatively, maybe use Pick's Theorem. Pick's Theorem states that for a simple polygon with integer coordinates, the area A is equal to I + B/2 - 1, where I is the number of interior lattice points and B is the number of boundary lattice points. However, in our case, the square doesn't necessarily have vertices at lattice points, so Pick's Theorem might not apply directly. But maybe we can estimate?But the problem is asking for the maximum number of lattice points covered, not necessarily the number of interior or boundary points. So maybe Pick's Theorem isn't directly helpful here.Wait, another thought: the maximum number of lattice points a square can cover is related to how many grid lines it intersects. Each time the square crosses a vertical or horizontal grid line, it potentially enters a new column or row, which could contain a lattice point. So, the number of vertical grid lines intersected is roughly the ceiling of the side length, and similarly for horizontal. But if the square is rotated, it might intersect more grid lines.But maybe the number of vertical grid lines crossed by a square is related to the length of its projection onto the x-axis. Similarly for horizontal. If the square is axis-aligned, then the number of vertical grid lines it crosses is floor(s) + 1, which would be n + 1 for s = n + ε. If it's rotated, the projection is longer, so it might cross more vertical grid lines.For a square rotated by θ, the projection onto the x-axis is s*|cosθ| + s*|sinθ|, assuming the square is rotated such that both directions contribute. Wait, actually, the width of the projection would be s*(cosθ + sinθ) if it's rotated by θ from the axis. So, to maximize the projection, we set θ=45 degrees, giving s*sqrt(2). For our case, s = n + 1/(2n+1). So, the maximum projection is s*sqrt(2). Therefore, the number of vertical grid lines crossed would be floor(s*sqrt(2)) + 1? Wait, not exactly. Because the projection is the width, so the number of vertical lines crossed is the number of integers between the left and right boundaries of the square.So, if the square's projection on the x-axis spans from a to b, then the number of vertical grid lines crossed is floor(b) - ceil(a) + 1. Similarly for the y-axis. Therefore, to maximize the number of lattice points, we need to maximize the number of vertical and horizontal grid lines crossed, which would be achieved by maximizing the span of the projection.But since we can translate the square anywhere, the maximum number of vertical grid lines crossed would be floor(sqrt(2)*s). Wait, but actually, the projection length is s*(cosθ + sinθ). So, if we rotate the square to maximize the projection, which is at θ=45 degrees, the projection length is s*sqrt(2). Therefore, the maximum number of vertical grid lines crossed would be floor(s*sqrt(2)) + 1. Similarly for horizontal. Therefore, the number of lattice points would be [floor(s*sqrt(2)) + 1]^2.But s here is n + 1/(2n+1). Let's compute s*sqrt(2):s*sqrt(2) = (n + 1/(2n+1)) * sqrt(2).But 1/(2n+1) is a small term. Let me see:For example, n=1:s*sqrt(2) = (1 + 1/3)*sqrt(2) ≈ 1.333*1.414 ≈ 1.885. So floor(1.885) +1 =1 +1=2. So [2]^2=4. Which matches the earlier result.For n=2:s*sqrt(2) = (2 + 1/5)*sqrt(2) ≈ 2.2*1.414 ≈ 3.11. Then floor(3.11) +1=3 +1=4. So [4]^2=16. Hmm, but if the axis-aligned square for n=2 would have side length 2 + 1/5 = 2.2, and if it's axis-aligned, the number of lattice points would be floor(2.2) +1=2 +1=3 in each direction, so 3x3=9. But the rotated square could cover 4x4=16? That seems like a huge difference, but maybe it's possible?Wait, but if the square is rotated, does it actually cover all those lattice points? For example, if we have a square of side length 2.2 rotated at 45 degrees, covering a projection of ~3.11 units, but the actual square is diamond-shaped. Each point in the grid would be covered if it lies within the square. However, even though the projection is 3.11, the square itself is smaller. So maybe the number of lattice points isn't actually as high as 16. There must be some overestimation here.Alternatively, perhaps the formula [floor(s*sqrt(2)) +1]^2 is an upper bound, but the actual number might be less. Because the square's area is s^2, and the number of lattice points can't exceed the area by too much. For s=2.2, area is ~4.84, but 16 lattice points would mean each lattice point has only 0.3 area, which is impossible. Therefore, my previous reasoning is flawed.Wait, so maybe the key is that when the square is rotated, even though its projection is longer, the actual number of lattice points it can cover is constrained by its area. So the maximum number of lattice points is roughly proportional to the area of the square plus some term related to the perimeter.In that case, using the formula from the geometry of numbers: the number of lattice points in a convex body K is approximately Area(K) + O(Perimeter(K)). But I need to check the exact theorem.According to the Gauss circle problem, for a circle, the error term is O(r), but for general convex sets, the error term can be proportional to the perimeter. So, perhaps for a square, the number of lattice points is approximately Area + O(Perimeter). Therefore, for our square of side length s, the area is s², and the perimeter is 4s. Therefore, the number of lattice points is s² + O(s). Since s = n + 1/(2n+1) ≈ n, the leading term would be n², and the error term would be O(n). But the problem is asking for the exact maximum number, not an asymptotic estimate.Wait, but given that the side length is n + 1/(2n+1), which is slightly larger than n, the area is (n + 1/(2n+1))² = n² + 2n/(2n+1) + 1/(2n+1)². So approximately n² + 1 + something small. So the area is about n² +1. Therefore, the number of lattice points can't be much higher than n² +1. But when n=1, the area is (4/3)^2 ≈ 1.78, and the maximum number of lattice points is 4, which is higher than the area. For n=2, area ≈ 4.84, and maybe 9 lattice points? Which is again higher.Wait, so area is not a strict upper bound. Lattice points can be more than the area because each lattice point is a unit area, but overlapping counts. So, for example, a square can cover multiple lattice points even if its area is less than the number of points.But then, how do we find the maximum?Wait, perhaps looking at the problem from another angle. Suppose we have a square of side length n + ε, where ε is very small. Then, the square can be positioned to just barely cover an extra row and column beyond n, but due to the small ε, maybe not. However, in our case, ε is 1/(2n+1), which is a specific value.Wait, 1/(2n+1) is the reciprocal of an odd integer. For example, when n=1, ε=1/3, which is 0.333... When n=2, ε=1/5=0.2. So, as n increases, ε decreases.Maybe the key is to position the square such that it's centered between lattice points, with its sides aligned in such a way that it can cover as many points as possible. For example, if the square is centered at (0.5, 0.5), then its sides are between lattice points, allowing it to cover points around it.Alternatively, consider that the square's side length is n + 1/(2n+1). Let's write this as n + δ, where δ=1/(2n+1). So, the total side length is n + δ. If we position the square such that it starts at (0,0), it would extend to (n + δ, n + δ). The number of lattice points in this axis-aligned square would be (n +1)^2, since x and y can range from 0 to n inclusive. But if we shift the square slightly, maybe we can include an extra column or row.Wait, for example, if we shift the square so that it starts at (-δ/2, -δ/2), then it would extend from (-δ/2, -δ/2) to (n + δ/2, n + δ/2). The length in each direction is still n + δ. Now, the x-coordinates covered would be from -δ/2 to n + δ/2. The integer x-coordinates within this interval would be 0, 1, ..., n. Similarly for y. So again, (n+1)^2 points. So shifting the square doesn't help here.But maybe if we rotate the square by some angle so that it can cover more lattice points. For example, if we tilt the square such that its sides cross through multiple lattice points.Alternatively, use the concept of the smallest square that can cover (m+1)x(m+1) lattice points. The side length required for a square to cover (m+1)x(m+1) lattice points when axis-aligned is m, but if rotated, perhaps it can be smaller. However, in our problem, it's the opposite: given the side length, find the maximum number of lattice points.Wait, another approach: the maximum number of lattice points is the number of integer points (i,j) such that the square [a, a + s] x [b, b + s] contains (i,j). To maximize this, we need to choose a and b such that as many (i,j) as possible are inside the square.But since the square can be placed anywhere, the maximum number is the same as the maximum number of integer points that can fit in a square of side length s. For axis-aligned squares, this is (floor(s) +1)^2. But if we rotate the square, perhaps we can fit more.However, I remember that the minimal area required to cover k lattice points is roughly k - something, but I don't recall the exact result. Maybe we need a different strategy.Wait, let's consider the Minkowski's theorem or some other geometry of numbers theorem, but I think they are more about lattices and convex bodies, not directly applicable here.Alternatively, think about the problem in terms of how the square can be placed to cover the most lattice points. Since the square has a side length just a bit larger than n, perhaps the optimal configuration is when the square is placed such that it covers an n x n grid of points, plus an additional row and column, making it (n+1)x(n+1). But the side length required for an axis-aligned square to cover (n+1)x(n+1) points is n, since from 0 to n inclusive. But our square has a side length larger than n, so it can definitely cover (n+1)x(n+1) points. But can it cover more?Wait, for example, if n=1, side length 4/3. As we saw, it can cover 4 points. If we make the square axis-aligned, but shift it slightly, can it cover 5 points? Let's see. Suppose the square is placed from 0.1 to 1.433 in both x and y. Then, the integer points inside would be (0,0), (0,1), (1,0), (1,1). Still 4 points. Even if you shift it, as long as the square's sides are between 0 and 1.433, you can't include x=2 or y=2. So 4 points.But if we rotate the square, maybe we can include a fifth point? For example, if the square is rotated so that its corners are near lattice points. Wait, if the square is rotated by 45 degrees and scaled, but our side length is fixed. Let me calculate.Suppose we have a square of side length 4/3 rotated by 45 degrees. The distance between two adjacent lattice points diagonally is sqrt(2). The diagonal of the square is (4/3)*sqrt(2) ≈ 1.885. So, the diagonal is less than 2 units. Therefore, the rotated square's diagonal is less than 2, meaning it can't reach from one lattice point to the next diagonally. So, perhaps it can cover the center point and four surrounding points? Let me visualize.If the square is centered at (0.5, 0.5) and rotated by 45 degrees, with diagonal ~1.885, which would mean each side is 4/3. The vertices of the square would be at (0.5 ± 2/3, 0.5 ± 2/3). So, from approximately (0.5 - 0.666, 0.5 - 0.666) ≈ (-0.166, -0.166) to (0.5 + 0.666, 0.5 + 0.666) ≈ (1.166, 1.166). So, the integer points inside would be (0,0), (0,1), (1,0), (1,1). Again, 4 points. So, rotating doesn't help here.Hmm. So for n=1, the maximum is 4. For n=2, the side length is 2 + 1/5 = 11/5 = 2.2. If axis-aligned, from 0 to 2.2, covering x=0,1,2 and y=0,1,2, so 9 points. If rotated, the diagonal is 2.2*sqrt(2) ≈ 3.11. If centered at (1.1, 1.1), rotated, the square would extend from approximately 1.1 - 1.555 ≈ -0.455 to 1.1 + 1.555 ≈ 2.655 in both x and y. So, integer points from -0 to 2, so x=0,1,2 and y=0,1,2, giving 9 points. But maybe if we shift the rotated square, we can cover more?Wait, if we shift the rotated square to the right, maybe cover x=0,1,2,3? But the projection is ~3.11, so if we start the square near x=0, the projection would go to x≈3.11. But since the square's actual coverage in x is from a to a + 3.11, but rotated, the actual square's position might not cover all those x-values. Wait, this is getting confusing.Alternatively, maybe there's a mathematical formula or existing result about the maximum number of lattice points a square of given side length can cover. Let me try to recall or derive it.Suppose the square is rotated by an angle θ and has side length s. The number of lattice points it can cover depends on how it is positioned relative to the grid. To maximize the number of lattice points, the square should be positioned such that as many lattice points as possible lie inside it.An approach could be to model the square as a rotated rectangle and use the rotation angle to calculate coverage. However, this might be complex.Alternatively, consider that the maximum number of lattice points occurs when the square is aligned such that its sides are mid-way between lattice points, thereby maximizing the number of rows and columns it can intersect. But I need to formalize this idea.Wait, perhaps using the concept of the minimal number of lattice points that can define a square, but not sure.Wait, here's another idea: the number of lattice points inside a square is maximized when the square is positioned such that its sides are as close as possible to lattice lines without excluding any potential lattice points. For a square of side length slightly larger than n, aligning it to cover n+1 columns and rows would be optimal. However, since the side length is n + ε, you can't cover n+1 full units unless ε is sufficient.But in our case, ε is 1/(2n+1). Let's compute n + 1/(2n+1). For example, n=1: 1 + 1/3 = 4/3 ≈1.333. So, in x-direction, the square spans 1.333 units. If aligned from 0 to 1.333, it includes x=0,1 but not 2. Similarly, if shifted, maybe from 0.333 to 1.666, which would still include x=1 but not 2. So, in any case, only two x-coordinates. Therefore, axis-aligned gives (n+1)^2=4 points.Wait, but when n increases, say n=2, side length=2 + 1/5=2.2. If axis-aligned, covers x=0,1,2. If shifted a bit, from 0.1 to 2.3, still x=0,1,2. So 3 x-coordinates. So 3x3=9 points. If rotated, can we cover more?The diagonal of the square is 2.2*sqrt(2)≈3.11. If the square is rotated by 45 degrees and centered appropriately, maybe covering x=0,1,2,3?Wait, let's suppose the square is rotated and centered at (1.5,1.5). Then, the projection on the x-axis would be from 1.5 - 3.11/2 ≈1.5 -1.555≈-0.055 to 1.5 +1.555≈3.055. So, x-coordinates from -0.055 to 3.055, which includes x=0,1,2,3. Similarly for y. So, total lattice points would be 4x4=16. But the square has area 2.2²=4.84, which is way less than 16. This is impossible. Therefore, my assumption is wrong. A square of area ~4.84 cannot cover 16 lattice points.Wait, but the problem is the rotated square's area is the same as the original square. So, the area is (n + 1/(2n+1))². If the square is rotated, it's the same area. But how can a square of area ~4.84 cover 16 lattice points? Each lattice point is spaced 1 unit apart, so the density is 1 point per unit area. 4.84 area can't have 16 points. So, clearly, my previous reasoning about the projections leading to 16 points is flawed.Therefore, the number of lattice points must be related to the area, but with some multiplier. However, in reality, the maximum number of lattice points a square can cover is limited by both its area and its arrangement. So, perhaps the correct approach is to use the axis-aligned square, which gives (n+1)^2 points, and see if rotation allows us to cover more.But for the rotated square, even though the projection is longer, the actual coverage of lattice points isn't as straightforward. For example, a square rotated by 45 degrees will have its corners in between lattice points, but the sides may pass through several lattice points.Wait, let's think of a diamond-shaped square (rotated 45 degrees) with side length sqrt(2)*k, which would have vertices at (k,0), (0,k), (-k,0), (0,-k). Such a square would cover lattice points where |x| + |y| <= k. But in our case, the square has side length n + 1/(2n+1). Wait, if we rotate it by 45 degrees, the "radius" (distance from center to vertices) would be (n + 1/(2n+1)) * sqrt(2)/2. Let's compute that.For example, n=1: radius = (4/3)*sqrt(2)/2 ≈ 0.9428. So, lattice points with |x| + |y| <= 0.9428. But |x| + |y| must be integer or half-integer. The maximum |x| + |y| would be 0, so only the origin? But that contradicts earlier results. Wait, maybe my coordinate system is off.Alternatively, perhaps the rotated square is placed such that its sides are between the grid points. For example, centered at (0.5, 0.5), the rotated square with side length 4/3 would have vertices at (0.5 + 2/3, 0.5), (0.5, 0.5 + 2/3), (0.5 - 2/3, 0.5), (0.5, 0.5 - 2/3). This forms a diamond shape. The lattice points inside would be those (x,y) such that the distance from (0.5,0.5) is less than or equal to 2/3 in both x and y directions? No, actually, the diamond is defined by |x - 0.5| + |y - 0.5| <= 2/3. So, solving this inequality: the maximum |x - 0.5| + |y - 0.5| is 2/3. For lattice points (integers x,y), the left-hand side becomes |x - 0.5| + |y - 0.5|. For x and y integers, this is 0.5 + 0.5 =1 for adjacent points, which is more than 2/3 ≈0.666. Therefore, no lattice points would be inside this square. Wait, that can't be. Wait, when I earlier considered n=1, the rotated square did cover four points. There's a contradiction here.Wait, perhaps my mental model is incorrect. Let me get back to basics.A square rotated by 45 degrees with side length s will have its vertices at a distance of s*sqrt(2)/2 from the center. So, if the center is at (0.5, 0.5), the vertices are at (0.5 ± s*sqrt(2)/2, 0.5) and (0.5, 0.5 ± s*sqrt(2)/2). For s=4/3, this distance is (4/3)*sqrt(2)/2 ≈0.9428. So, the vertices are at approximately (0.5 + 0.9428, 0.5) ≈(1.4428, 0.5), and similar for others. Therefore, the square extends from approximately -0.4428 to 1.4428 in both x and y directions. Therefore, the integer x and y coordinates covered are 0 and 1. So, lattice points (0,0), (0,1), (1,0), (1,1). So, four points, same as axis-aligned.Therefore, for n=1, rotated or not, the number of lattice points covered is four. Which is (1+1)^2.For n=2, side length=11/5=2.2. Rotated square's vertices would be at a distance of 2.2*sqrt(2)/2≈1.555 from the center. If centered at (1.1,1.1), the square would extend from 1.1 -1.555≈-0.455 to 1.1 +1.555≈2.655 in x and y. Therefore, integer coordinates from 0 to 2, inclusive. So, 3x3=9 points. Which is (2+1)^2=9.Wait, so it's the same as the axis-aligned case. Therefore, rotating the square doesn't actually increase the number of lattice points covered. That seems counterintuitive, but perhaps it's correct.Wait, but if the side length is increased beyond n +1, then rotation would help. For example, a square of side length sqrt(2) (~1.414) rotated by 45 degrees can cover 5 lattice points: (0,0), (1,0), (0,1), (-1,0), (0,-1). But an axis-aligned square of side length 1.414 would cover 4 points. So, rotation helps there. But in our problem, the side length is n + 1/(2n+1), which is less than n +1. So, maybe rotation doesn't help here.Therefore, perhaps the maximum number of lattice points a square of side length n + 1/(2n+1) can cover is (n+1)^2, same as the axis-aligned case.But let's test with n=1: side length=4/3, axis-aligned covers 4 points. Any rotated or shifted square also covers 4 points. So, yes.For n=2: side length=11/5=2.2, axis-aligned covers 9 points. If I rotate the square, can I cover more? Let's see.Suppose the square is rotated by θ, not necessarily 45 degrees. Maybe a different angle. For example, angle such that the square's sides cross through multiple lattice points.Alternatively, use the following approach: the maximum number of lattice points is achieved when the square is positioned such that as many of its edges as possible pass through lattice points, thereby including them.But in order for a lattice point to be on the edge of the square, the square's side must align with the grid or be at an angle that causes the edge to intersect the lattice point.Alternatively, consider that the number of lattice points on the perimeter of the square can be maximized by aligning edges with rational slopes, but this is getting too vague.Wait, perhaps another strategy: the side length is n + 1/(2n+1). Let's denote s = n + ε, where ε =1/(2n+1). We can write ε as 1/(2n+1). So, s = n + 1/(2n+1).If we consider the continued fraction or some approximation, but not sure.Wait, notice that 2n+1 is the denominator in ε. Maybe there's a relation to Farey sequences or something, but I'm not sure.Alternatively, think of the square as covering a certain number of rows and columns. For the square to cover m rows and columns, the side length must be at least m -1. So, if the side length is s, then m -1 ≤ s < m. But in our case, s =n + 1/(2n+1). So, n < s < n +1. Therefore, the square can cover n +1 rows and columns if placed axis-aligned, but since s <n +1, you actually can't cover n +1 full units. Wait, no. If the square is axis-aligned, the number of rows and columns it can cover is floor(s) +1. Since s =n + ε, floor(s) =n, so floor(s) +1 =n +1. Therefore, it can cover n +1 rows and columns, hence (n +1)^2 lattice points.But even though the side length is less than n +1, the number of lattice points covered is still (n +1)^2. For example, a square from 0 to 0.5 in both axes covers 1 lattice point (0,0), which is (floor(0.5) +1)^2=1. Similarly, from 0 to 1.9, covers (1 +1)^2=4 points. So, the formula holds even when the side length is not an integer. Therefore, the axis-aligned square will always give (n +1)^2 lattice points, regardless of the fractional part of the side length, as long as it's greater than n.Therefore, if we place the square axis-aligned, starting at (0,0), it will cover all lattice points (i,j) where 0 ≤i ≤n and 0 ≤j ≤n, totaling (n +1)^2 points.But wait, the side length is s =n +1/(2n+1). So, the square extends from 0 to s in x and y. The maximum i such that i ≤s is floor(s) =n, since s <n +1. Hence, yes, lattice points are from 0 to n, inclusive. So, (n +1)^2.But can we place the square in a different position to cover more lattice points? For example, if we shift the square so that part of it is in negative coordinates.Suppose we center the square at (n/2, n/2). Then, the square would extend from (n/2 - s/2, n/2 - s/2) to (n/2 + s/2, n/2 + s/2). Let's compute the bounds.s =n +1/(2n+1). Therefore, s/2 =n/2 +1/(2*(2n+1)).The lower bound in x is n/2 - s/2 =n/2 - [n/2 +1/(2*(2n+1))] =-1/(2*(2n+1)).Similarly, upper bound in x is n/2 + s/2 =n/2 + [n/2 +1/(2*(2n+1))] =n +1/(2*(2n+1)).Therefore, the x-coordinates range from -1/(2*(2n+1)) to n +1/(2*(2n+1)). Since 1/(2*(2n+1)) is a positive number less than 1/2, the integer x-coordinates covered are from 0 to n, inclusive. Similarly for y. Therefore, the number of lattice points is still (n +1)^2.Therefore, centering the square doesn't help. Shifting it into negative coordinates doesn't include more points since the negative side is less than 1/2 unit, which isn't enough to include x=-1.Alternatively, if we shift the square by almost 1 unit in some direction. For example, shifting it so that it starts at (-1 + ε, -1 + ε), then it would extend to (-1 + ε + s, -1 + ε + s). If s =n +1/(2n+1), then the upper bound would be (-1 + ε +n +1/(2n+1)). If we choose ε such that -1 + ε +n +1/(2n+1) >=n, then the upper bound is >=n, which would still cover x from -1 to n-1, but wait, no.Wait, let me compute:If we shift the square to start at x = a, then it ends at x = a + s. To cover the lattice points from x = k to x = m, we need a ≤k ≤m ≤a + s.If we set a = -1 + ε, then the upper bound is -1 + ε + s = -1 + ε +n +1/(2n+1). For this to be ≥n, we need -1 + ε +n +1/(2n+1) ≥n ⇒ -1 + ε +1/(2n+1) ≥0 ⇒ ε ≥1 -1/(2n+1). But ε is a small positive number (shift amount). If we set ε=1 -1/(2n+1), then a=-1 +1 -1/(2n+1)= -1/(2n+1). Then the square would extend from -1/(2n+1) to -1/(2n+1) +n +1/(2n+1)=n. Therefore, the x-coordinates covered are from -1/(2n+1) to n, which includes x=0 to x=n, same as before. So, no gain.Alternatively, shift further. But shifting into negative coordinates beyond 1 unit would require the square to have a larger side length to reach positive coordinates, which we don't have.Therefore, regardless of shifting, the number of lattice points covered axis-aligned is (n +1)^2.But what if we don't align the square with the axis? For example, rotate it so that its sides are not parallel to the axes. As I saw earlier, for n=1 and n=2, rotating the square doesn't increase the number of lattice points covered. But maybe for higher n?Wait, let's take n=3. Then, side length s=3 +1/7≈3.142857. An axis-aligned square would cover (3+1)^2=16 points. If we rotate the square by 45 degrees, the projection onto the x-axis is s*sqrt(2)≈3.142857*1.414≈4.45. So, the projection spans approximately4.45 units. If centered at (1.5,1.5), the square would extend from1.5 -4.45/2≈1.5 -2.225≈-0.725 to1.5 +2.225≈3.725. Therefore, integer x from0 to3, so4 values. Same for y. Therefore, 4x4=16 points. Which is the same as axis-aligned.But the area of the square is s²≈9.877, which is less than 16, so this suggests that the actual number of lattice points inside can't be 16. Contradiction again. Therefore, my method of counting lattice points via projections is incorrect.Therefore, I must conclude that rotating the square doesn't actually allow covering more lattice points than the axis-aligned case, at least for the side lengths given. Therefore, the maximum number of lattice points is indeed (n +1)^2.But wait, the problem says "determine the maximum number of lattice points in the plane a square of side length n +1/(2n+1) may cover." If the answer is always (n +1)^2, regardless of the rotation, then that's the answer. But is there a way to cover more?Wait, let's think differently. Suppose the square is placed such that each side is tangent to a lattice point. For example, if the square is positioned so that its sides just touch certain lattice points, thereby maximizing the number of points inside.Alternatively, arrange the square so that multiple lattice points lie on its boundary. Each side of the square can contain multiple lattice points if aligned correctly.However, given that the side length is n +1/(2n+1), which is not an integer, it's unclear how to align the sides to pass through multiple lattice points.Alternatively, use the fact that the side length is n +1/(2n+1). Let's write this as (2n(n) +1)/(2n+1). Wait, n +1/(2n+1) = (n*(2n+1) +1)/(2n+1) = (2n² +n +1)/(2n+1). Not sure if helpful.Alternatively, consider that 1/(2n+1) is a fraction that allows the square to just barely cover an extra lattice point when tilted. For example, if we tilt the square such that it aligns with a certain rational slope, the small addition of 1/(2n+1) allows the square to reach an extra lattice point.But I need to formalize this idea.Suppose we tilt the square at an angle θ such that tanθ = 1/k for some integer k. Then, the square's sides will have a slope that allows them to intersect lattice points. However, calculating this requires more precise analysis.Alternatively, consider that the square is positioned to have one side passing through (0,0) and (n,1), making the side length equal to sqrt(n² +1). But our side length is n +1/(2n+1), which is different.Wait, compute the difference between sqrt(n² +1) and n +1/(2n+1):sqrt(n² +1) ≈n +1/(2n) -1/(8n³) +... So, approximately, sqrt(n² +1) ≈n +1/(2n). Our side length is n +1/(2n+1). For large n, 1/(2n+1) ≈1/(2n), so they are similar. However, for finite n, 1/(2n+1) is slightly less than 1/(2n). So, sqrt(n² +1) is slightly larger than our side length. Therefore, a square with side length sqrt(n² +1) can be placed diagonally to cover (n+1)^2 lattice points, similar to an axis-aligned square. But our square is slightly smaller, so maybe it can't.But in our problem, the side length is n +1/(2n+1). If we can arrange the square to be diagonal, covering approximately the same area as an axis-aligned square but positioned to intersect more lattice points.Wait, another idea inspired by covering grid points with a square: the maximum number of lattice points is (n +1)^2 +1. For example, by positioning the square such that it includes an extra point outside the n x n grid. But with side length just over n, maybe it's possible.Wait, let's take n=1, side length=4/3. If I place the square from 0.333 to 1.666 in both x and y. Then, x=0.333 to1.666 includes x=0,1. Similarly for y. So, (0,0), (0,1), (1,0), (1,1). Still 4 points. But if I shift the square to include x=2? The side length is 4/3≈1.333. If I place the square from 0.666 to2.0, then it spans 1.333 units. The lattice points would be x=1,2. But since the square is shifted, it would cover x=1 and2, but in reality, x=2 is at 2.0, which is the edge of the square. Depending on whether the square is inclusive of its boundaries. If the square is closed, then x=2 is included. But lattice points are discrete, so (2, y) would be included if the square includes x=2. But if the square is placed from0.666 to2.0, then x=2 is included. So, lattice points x=1 and2. Similarly for y. So, points (1,1), (1,2), (2,1), (2,2). Also 4 points. So, no gain.But wait, if the square is placed from0.1 to1.433, does it include (1,1)? Yes, since1 is within0.1 to1.433. But (2,2) is outside. Therefore, no.Alternatively, place the square such that it covers two separate regions. But since the square is convex and a single connected shape, it can't cover non-contiguous lattice points.Therefore, perhaps the maximum number of lattice points is indeed (n +1)^2. But the problem states "a square of side length n +1/(2n+1)", so maybe there's a way to cover more points by cleverly rotating or positioning the square, leveraging the specific fraction 1/(2n+1).Wait, notice that 1/(2n+1) is a reciprocal of an odd integer. Maybe there's a way to relate this to the Farey sequence or a fraction that allows the square to align with some grid points when rotated.Alternatively, consider that the square can be positioned such that its sides are just long enough to cover an additional lattice point when rotated. For example, if a square is rotated such that its corners are near lattice points, the side length needed to reach from one lattice point to another diagonally is sqrt((a)^2 + (b)^2), where a and b are integers. If we can make the side length equal to such a diagonal, then the square can cover those lattice points at the corners.But in our case, the side length is n +1/(2n+1). Let's compute this for small n:n=1: 1 +1/3=4/3≈1.333. sqrt(1^2 +1^2)=sqrt(2)≈1.414. So, 4/3 is less than sqrt(2). Therefore, a square of side length 4/3 can't reach from (0,0) to (1,1). But it can reach from (0,0) to (1,0.333), for example.Alternatively, if we place the square such that two adjacent corners are at lattice points, and the other two are not. For example, from (0,0) to (1,0) to (1 + ε, ε) to (ε,1 + ε), forming a square. The side length would be sqrt(1^2 + ε^2). To have side length 4/3, set sqrt(1 + ε²)=4/3 ⇒ ε=sqrt((16/9) -1)=sqrt(7/9)=sqrt(7)/3≈0.882. So, the other two corners would be at (1 + sqrt(7)/3, sqrt(7)/3) and (sqrt(7)/3,1 + sqrt(7)/3). These points are not lattice points. Therefore, the square would have only two lattice points at (0,0) and (1,0), but the other points are non-integer. Therefore, not helpful.Alternatively, position the square so that multiple edges pass through lattice points. For example, if the square's edges are aligned to have a high number of lattice points on them. However, unless the side length is such that the edges can pass through multiple lattice points (e.g., a square with side length 5/3 can have edges passing through (0,0), (5/3,0), etc.), but again, not sure.Given all these considerations and examples where rotating the square doesn't increase the number of lattice points beyond (n +1)^2, I'm leaning towards the conclusion that the maximum number of lattice points is indeed (n +1)^2. However, the problem asks for the maximum number a square of side length n +1/(2n+1) may cover. Since the side length is greater than n, but less than n +1, it's possible that by rotating the square, we can cover more points than the axis-aligned case.Wait, let's recall a classic result: the number of lattice points in a square of side length s is at most (floor(s) +1)^2. But this is for axis-aligned squares. For rotated squares, the number can be higher. For example, a square rotated at 45 degrees with side length sqrt(2) can cover 5 lattice points, while an axis-aligned square of side length 1 covers 4.But in our problem, the side length is n +1/(2n+1). For large n, this approaches n +1/(2n). So, the question is, can we arrange this square such that it covers more than (n +1)^2 lattice points?Alternatively, think of the square as a rotated grid-aligned square plus some extra. For example, a square of side length n + ε can be thought of as a square that covers an n x n grid and has ε extra length to possibly cover an additional point.But how?Wait, maybe by shearing the square or arranging it such that the extra length allows the square to cover an extra diagonal point.For example, consider n=1, s=4/3. If we rotate the square such that it is positioned between two lattice points diagonally, the extra 1/3 length might allow it to cover an additional point. But as seen earlier, it doesn't.Wait, but here's a different approach. The side length is n +1/(2n+1). Let's compute the value of 2n+1:For a given n, the denominator is 2n+1. Let's denote m=2n+1. Then, the side length is n +1/m. Since m=2n+1, this can be rewritten as (n*m +1)/m = (n*(2n+1)+1)/(2n+1) = (2n² +n +1)/(2n+1).Not sure if that helps, but maybe there's a pattern here.Alternatively, consider that the excess length 1/(2n+1) allows the square to reach just enough to cover an additional lattice point when aligned in a specific way. For example, imagine a square that covers an n x n grid and has 1/(2n+1) extra length to reach an additional point in a particular direction.But how?Alternatively, place the square so that it covers a grid from (0,0) to (n,n), and use the extra length to include one more point in a row or column. However, since the side length is less than n +1, you can't fully include an additional column or row.But perhaps if the square is rotated such that the extra length allows it to cover an additional diagonal point.For example, a square of side length n + ε rotated such that it can reach from (0,0) to (n,n) plus an extra ε in some direction. But this is vague.Wait, let's think in terms of vectors. Suppose we have a square with one side vector (a,b) and the other side vector (-b,a) to keep it a square. The side length is sqrt(a² +b²)=n +1/(2n+1). We need to choose integers a and b such that sqrt(a² +b²)=n +1/(2n+1). However, since n +1/(2n+1) is not an integer or a rational number with a nice sqrt, this approach might not work.Alternatively, approximate the side length with a vector close to (n,0) but adjusted with a small component in another direction. But this is getting too abstract.Alternatively, the problem might be a direct application of a known result in combinatorial geometry. Perhaps the maximum number of lattice points is (n +1)^2 +n, but I need to verify.Wait, for example, take n=1: 4 +1=5. But as we saw, the maximum is 4. So that can't be.Alternatively, for n=1, maybe 5 points? But I don't see how.Wait, another approach: the number of lattice points covered by a square is related to the number of unit squares it overlaps. Each unit square can contribute at most one lattice point (the lower-left corner, for example). So, the number of lattice points is at most the number of unit squares intersected by the square. The number of unit squares a square of side length s intersects is roughly (s +1)^2, which for s =n +1/(2n+1) is (n +1/(2n+1) +1)^2≈(n +1)^2. But this is just a rough estimate.However, this is similar to the axis-aligned case. So, again, this suggests that the maximum number is (n +1)^2.Given all these considerations, and especially the examples for n=1 and n=2 where rotating the square doesn't increase the number of lattice points covered, I think the answer is (n +1)^2. Therefore, the maximum number of lattice points a square of side length n +1/(2n+1) can cover is (n +1)^2.But just to confirm, let's consider n=3. Side length=3 +1/7≈3.1428. Axis-aligned covers 16 points. If we rotate the square by a small angle, would it cover more? The projection on the x-axis would be s*sqrt(2)≈3.1428*1.414≈4.45. If centered at (1.5,1.5), the square extends from≈-0.725 to≈3.725, covering x=0,1,2,3. So 4x4=16 points. Same as axis-aligned. But the area is≈9.88, and 16 lattice points. Since each lattice point is spaced 1 unit apart, the density is 16/9.88≈1.62 points per unit area, which is feasible? Not sure. But since the square can overlap multiple unit squares, it can indeed cover multiple lattice points.However, in reality, a rotated square of area ~10 can cover 16 lattice points if its projection is such, but I need to visualize. If the square is rotated and spans from x≈-0.7 to x≈3.7, then yes, it covers 4 columns of x, each column contributing 4 rows. But if the square is diamond-shaped, it's possible that not all these points are inside. For example, the points at the corners of the grid (0,0), (3,3) might be near the rotated square's boundaries.But since the square's side length is ~3.14, the distance from the center to the vertices is ~1.555, which for a center at (1.5,1.5), the square reaches from≈-0.055 to≈3.055. So, lattice points from0 to3, inclusive. So, 4x4=16 points. Whether all these points are inside the rotated square depends on the square's position. If the square is rotated and centered at (1.5,1.5), then the distance from the center to any lattice point (i,j) is sqrt((i -1.5)^2 + (j -1.5)^2). For i,j from0 to3, the maximum distance is sqrt((2.5)^2 + (2.5)^2)=sqrt(12.5)=~3.54. But the square's vertices are only ~1.555 from the center, so the maximum distance from the center to any vertex is≈1.555. Therefore, lattice points (i,j) must satisfy sqrt((i -1.5)^2 + (j -1.5)^2) ≤1.555. Let's compute for (0,0): sqrt(2.25 +2.25)=sqrt(4.5)=~2.121 >1.555. So, (0,0) is outside the square. Similarly, (3,3) is also outside. Therefore, the rotated square centered at (1.5,1.5) with side length ~3.14 does NOT actually cover all 16 lattice points from0-3 in x and y. In reality, only the points closer to the center are inside.So, how many lattice points does it cover?Compute for each lattice point (i,j) where0 ≤i,j ≤3:Distance from (1.5,1.5) is sqrt((i -1.5)^2 + (j -1.5)^2).For (1,1): sqrt(0.5^2 +0.5^2)=sqrt(0.5)=~0.707 <1.555. Inside.For (1,2): same as above. Inside.Similarly, (2,1), (2,2): Inside.For (0,1): sqrt(1.5^2 +0.5^2)=sqrt(2.25 +0.25)=sqrt(2.5)=~1.581 >1.555. Outside.Similarly, (1,0), (2,3), (3,2): same distance. Outside.For (1,3): sqrt(1.5^2 +1.5^2)=sqrt(4.5)=~2.121. Outside.Similarly for others.Therefore, only the four points around the center are inside: (1,1), (1,2), (2,1), (2,2). Plus the center point if it's a lattice point, but (1.5,1.5) isn't. So, only 4 points. Wait, that contradicts the projection idea. So, in reality, the rotated square of side length ~3.14 covers only 4 lattice points, much fewer than the axis-aligned case.Therefore, my previous reasoning about the projections was entirely incorrect. The number of lattice points inside a rotated square isn't simply determined by the projection span but by the actual positions relative to the square.Therefore, the maximum number of lattice points is indeed achieved by the axis-aligned square, giving (n +1)^2 points.Hence, after careful consideration and multiple examples, I conclude that the maximum number of lattice points a square of side length n +1/(2n+1) can cover is (n +1)^2.Final AnswerThe maximum number of lattice points is boxed{(n + 1)^2}.</think>Given a positive integer ( n ), we need to determine the maximum number of lattice points a square of side length ( n + frac{1}{2n+1} ) may cover.1. Lattice Points Definition: Lattice points are points with integer coordinates.2. Side Length Analysis: The side length of the square is ( n + frac{1}{2n+1} ), which is slightly more than ( n ) but less than ( n + 1 ).3. Axis-Aligned Square: If the square is axis-aligned, it spans from an initial point (e.g., ( 0 )) to ( n + frac{1}{2n+1} ). The number of lattice points covered in this case is determined by the integer coordinates from ( 0 ) to ( n ) inclusive, resulting in ( (n + 1)^2 ) lattice points.4. Rotated Square Consideration: Rotating the square might seem beneficial, but the projections and actual coverage were analyzed. For example, rotating the square by 45 degrees and checking the coverage showed that the number of lattice points did not exceed ( (n + 1)^2 ).5. Examples and Verification: Specific examples for ( n = 1 ), ( n = 2 ), and ( n = 3 ) were checked. In each case, the axis-aligned square covered ( (n + 1)^2 ) lattice points, while rotated squares either covered the same or fewer points due to their actual position relative to lattice points.6. Conclusion: The key insight is that the side length ( n + frac{1}{2n+1} ) is sufficient to cover ( n + 1 ) rows and columns when axis-aligned, but not enough to cover more when rotated or shifted.Thus, the maximum number of lattice points a square of side length ( n + frac{1}{2n+1} ) may cover is (boxed{(n + 1)^2}).,<think>Okay, so I have this problem here where I need to find the maximum number of lattice points a square with side length n + 1/(2n+1) can cover. The square is in the plane, and n is a positive integer. Hmm, let me think about how to approach this.First, lattice points are points with integer coordinates, right? So, I need to figure out how to place a square of that specific size so that it covers as many of these integer grid points as possible. The key here is probably figuring out the optimal orientation and position of the square to maximize the count.I remember that the number of lattice points a shape covers is related to its area and its perimeter, but I think that's more for when you're estimating using something like Pick's Theorem. But in this case, since it's a square, maybe the orientation can affect the number of points it can cover. For example, a square aligned with the grid (axis-aligned) will intersect different points compared to a square that's rotated.Wait, but when you rotate a square, you might be able to cover more points because the sides can pass through more lattice points? Or maybe less? I need to recall how lattice points are distributed with respect to lines. A line from (0,0) to (n, n) passes through certain lattice points, but a line that's at a different angle might pass through more or fewer.But perhaps the maximum number of lattice points covered by a square is achieved when the square is axis-aligned. Let me check that intuition. If the square is axis-aligned, then its sides are parallel to the x and y axes, so the number of lattice points it covers is basically the number of integer coordinates within its bounds. For a square of side length s, the number of lattice points would be roughly (floor(s) + 1)^2, but since the side length here is n + 1/(2n+1), which is slightly larger than n, maybe the number is (n + 1)^2? But wait, if the side length is n + 1/(2n+1), which is less than n + 1, then an axis-aligned square might only cover (n + 1)^2 points if it's placed such that it spans from 0 to n + 1/(2n+1) in both x and y. But since the side length is just over n, maybe it can cover (n + 1)^2 points if positioned correctly?Wait, hold on. If the side length is n + ε, where ε is a small positive number (in this case, ε = 1/(2n+1)), then the square can cover an area that's just slightly larger than n x n. An axis-aligned square of side length n will cover (n + 1)^2 lattice points, right? Because from x=0 to x=n, inclusive, there are n + 1 integers. Similarly for y. So, an n x n square placed from (0,0) to (n,n) covers (n+1)^2 points. But if we have a slightly larger square, say with side length n + ε, then if we shift the square to cover from x= -ε/2 to x= n + ε/2, then in each dimension, we might still only cover n + 1 points because the extra ε/2 on each side might not reach the next integer. However, if ε is large enough to include another integer, then it would cover more.But in our case, ε = 1/(2n+1), which is less than 1/(2n) since the denominator is 2n + 1. So, ε is quite small. Therefore, shifting the square by less than 1/(2n) on each side won't reach the next integer. Therefore, an axis-aligned square of side length n + 1/(2n+1) can be positioned such that it covers (n + 1)^2 lattice points, same as the n x n square. Wait, but the side length is actually longer, but the positioning is key. If it's shifted slightly, maybe it can cover an additional column or row?But to cover an additional column or row in the lattice, the square needs to span an additional integer in that direction. For example, if the square spans from x = a to x = a + n + 1/(2n+1). If a is chosen such that a + n + 1/(2n+1) >= k + 1 for some integer k, then the number of columns covered is (k + 1 - floor(a)). Hmm, this is getting a bit confusing. Maybe an example would help.Let me take n = 1. Then the side length is 1 + 1/(2*1 + 1) = 1 + 1/3 ≈ 1.333. If I place this square axis-aligned from x=0 to x≈1.333 and similarly for y, then the x-coordinates covered are 0 and 1 (since 1.333 doesn't reach 2). Similarly for y. So, the number of lattice points would be 2x2=4. But if I shift the square slightly, say from x=-0.1 to x=1.233, then it still covers x=0 and x=1. So, shifting doesn't help here. However, if I rotate the square, maybe it can cover more points?Wait, when you rotate a square, the projection onto the x and y axes becomes longer. So, a rotated square might span more integer coordinates along both axes. For example, a square rotated at 45 degrees with side length s has a diagonal of length s*sqrt(2). But the projections on the axes would be s*sqrt(2)/2. So, for a square of side length 1 + 1/3 ≈1.333, the projection would be about 1.333 * sqrt(2)/2 ≈0.9428. So, less than 1. Therefore, even if rotated, the projections might not span more than 1 unit, so it might not cover more lattice points. Hmm, maybe not.But wait, maybe there's another angle where the square can be rotated such that its sides pass through more lattice points, thereby covering more. But I need to think about how the rotation affects the number of grid points covered.Alternatively, maybe the maximum number of lattice points is indeed (n + 1)^2, achieved by an axis-aligned square, and the rotation doesn't help because the side length isn't enough to cover additional points. But I need to verify this for higher n.Let me take n = 2. Then side length is 2 + 1/(5) = 2.2. An axis-aligned square from x=0 to x=2.2 would cover x=0,1,2, so 3 columns. Similarly, 3 rows, so 9 points. If I rotate the square by some angle, would it cover more?The projection on each axis for a rotated square is side length * cos(theta) + side length * sin(theta), where theta is the angle between the side and the axis. Wait, actually, the width and height of a rotated square can be computed as s*|cos(theta)| + s*|sin(theta)|, since when you rotate a square, its bounding box is determined by the projections of the sides.But if the original side length is s, then the width and height of the rotated square (the bounding box) would be s*(cos(theta) + sin(theta)). To maximize the number of lattice points, we need the bounding box to cover as many integer coordinates as possible. So, the area would be roughly [s*(cos(theta) + sin(theta))]^2, but the number of lattice points would be roughly (floor(s*(cos(theta) + sin(theta))) + 1)^2.But we need to maximize this. The maximum value of cos(theta) + sin(theta) is sqrt(2), achieved at theta = 45 degrees. So, if we rotate the square by 45 degrees, the bounding box becomes s*sqrt(2). For n=2, s=2.2, so the bounding box would be 2.2*sqrt(2) ≈3.11. So, the bounding box would span about 3.11 units, which would cover 4 columns and 4 rows (since 3.11 is between 3 and 4). Therefore, the number of lattice points would be 4x4=16. But wait, the original axis-aligned square only covered 9 points. That seems contradictory. Wait, but the actual square is smaller, but its bounding box is larger. However, the square itself rotated would not cover all the points in the bounding box. So, maybe the number of points covered is actually less?Wait, no. When you rotate a square, even though the bounding box is larger, the square itself may pass through more lattice points. For example, a square rotated at 45 degrees can have its edges pass through more lattice points if placed correctly. But how does that affect the count?Alternatively, maybe using the rotated square allows the square to intersect more lattice points because the sides are now diagonal lines which can cross through more grid lines. For example, a line with a slope of 1 will pass through a lattice point every unit step. But a square's edge, if aligned diagonally, might pass through several lattice points along its edge.But the key here is whether the rotated square can cover more interior points. The area of the square is (n + 1/(2n+1))^2. By Pick's theorem, the area A is equal to I + B/2 - 1, where I is the number of interior lattice points and B is the number of boundary lattice points. But I don't know if Pick's theorem will directly help here since we need the total lattice points (interior plus boundary), which is I + B.But maybe not. Alternatively, perhaps we can estimate the maximum number of lattice points as roughly the area plus some term related to the perimeter. But the problem is asking for the exact maximum, not an estimate.Wait, but the problem says "determine the maximum number of lattice points in the plane a square of side length n + 1/(2n+1) may cover." So it's an exact value, depending on n.So maybe the maximum is indeed (n + 1)^2 + n^2? Wait, no, that seems too much. Let me think again.Alternatively, maybe when you rotate the square optimally, you can cover more lattice points. For example, if the square is rotated such that its sides are not aligned with the grid, then each side can potentially cut through more lattice points, thereby increasing the total count. But does this actually result in more lattice points inside the square?Wait, maybe the maximum number of lattice points is achieved when the square is positioned such that its sides pass through as many lattice points as possible. But how does that affect the count?Alternatively, perhaps the key is that the side length is specifically n + 1/(2n+1). This seems like a very particular choice. Maybe there's a way to position the square so that it just barely includes an extra column or row of lattice points compared to an axis-aligned square.Wait, let's analyze the axis-aligned case first. If the square is axis-aligned with side length n + 1/(2n+1), then in each dimension, the span is n + 1/(2n+1). Since 1/(2n+1) is less than 1, the number of integer coordinates covered in each dimension is from floor(a) to floor(a + n + 1/(2n+1)), where a is the starting coordinate. If we position the square such that it starts at x=0, then it goes up to x= n + 1/(2n+1). Since n is integer, the maximum x-coordinate covered is just a bit more than n, so the integer coordinates covered are 0,1,...,n, which is n + 1 points. Similarly in the y-direction. Therefore, the total number of lattice points is (n + 1)^2.But if we shift the square slightly, say by 1/(2n+1)/2 in both x and y directions, then the span becomes from -1/(2(2n+1)) to n + 1/(2n+1) - 1/(2(2n+1)) = n + 1/(2(2n+1)). But since 1/(2(2n+1)) is very small, less than 1/(2), so shifting by that amount still doesn't bring the square to include another integer in either direction. Therefore, the number of lattice points remains (n + 1)^2.But if we rotate the square, maybe we can cover more? Let's think. For example, when a square is rotated, its projections on the axes are longer, so perhaps it can cover more integer coordinates in each axis direction. However, the actual points covered might not be more because the square is not aligned with the grid.Alternatively, consider that rotating the square might allow it to intersect more lattice points along its edges. Each edge of the square can potentially pass through multiple lattice points if oriented correctly. If the square is rotated such that its edges have rational slopes, then the edges may pass through multiple lattice points. For example, a square rotated such that its sides have a slope of 1/1, 1/2, etc. However, the number of lattice points on each edge depends on the slope.But how does that affect the total count? If each edge passes through more lattice points, then the boundary of the square has more lattice points, but does that mean the interior also has more?Alternatively, perhaps the total number of lattice points covered by the square can be higher if the square is positioned such that it intersects many grid lines, thereby creating more regions with lattice points. However, I need to formalize this idea.Alternatively, think of the square as a union of rows and columns. If the square is rotated, each row and column might intersect the square in a different way, potentially allowing more lattice points to be inside.But I need a more systematic approach. Let's recall that for a square with side length s, the maximum number of lattice points it can cover is roughly (floor(s) + 1)^2 if axis-aligned. But if rotated, perhaps the number can be higher. However, when you rotate, even though the projections are longer, the actual points inside might not increase because the area is the same. Wait, the area is s^2. The number of lattice points is approximately the area plus some error term related to the perimeter.But the problem is to find the exact maximum. So perhaps there's a specific placement where the square can cover more than (n + 1)^2 points.Wait, let's calculate the area of the square: (n + 1/(2n+1))^2 = n^2 + 2n/(2n+1) + 1/(2n+1)^2. Since n is a positive integer, 2n/(2n+1) is slightly less than 1, so the total area is slightly less than n^2 + 1 + something negligible. Therefore, the area is approximately n^2 + 1. But by Pick's theorem, the number of lattice points I + B/2 - 1 = Area. So if we denote the total lattice points as I + B, then I + B = Area + B/2 + 1. Therefore, to maximize I + B, we need to maximize B, the number of boundary points.Therefore, if the square is positioned such that it has a large number of boundary lattice points, then the total number of lattice points would be higher. Therefore, a square that is rotated such that its edges pass through many lattice points would have a higher B, thereby increasing the total lattice points.So maybe the maximum number of lattice points is achieved when the square is rotated such that each side contains as many lattice points as possible. Let's think about how to arrange that.A line segment from (0,0) to (a,b) where a and b are integers will pass through gcd(a,b) + 1 lattice points. So, if we can make the sides of the square have a direction with a high gcd, then each side would pass through more lattice points.But the side length of the square is n + 1/(2n+1). If we can express the sides as vectors with integer components, then the number of lattice points on the side would be gcd(delta_x, delta_y) + 1. But since the side length is not an integer, but n + 1/(2n+1), which is not a rational number unless 1/(2n+1) is rational, which it is, but combined with n, it's n + 1/(2n+1). Hmm, but the vector representing the side would need to have length n + 1/(2n+1). If we can find a rational direction (a,b) such that sqrt(a^2 + b^2) = n + 1/(2n+1). But that seems difficult because sqrt(a^2 + b^2) is generally irrational unless a and b form a Pythagorean triple.Alternatively, maybe approximate such a direction. Wait, but this is getting too vague. Let me try a different approach.Suppose the square is rotated at an angle θ, then the number of lattice points it can cover depends on how the grid lines intersect the square. Maybe using the concept of the grid crossing number. The number of lattice points inside the square can be approximated by the area plus the number of intersections with the grid lines, but this is heuristic.Alternatively, consider that the maximum number of lattice points is achieved when the square is positioned such that its sides are between two lattice points, thereby maximizing the number of rows and columns it spans. Wait, but how?Wait, another idea: if the square is placed such that it is centered at a lattice point, then its sides are symmetric around that point. But I don't know if that helps.Alternatively, place the square such that its center is at the intersection of four lattice points, so that small shifts can cover different points. But since the side length is n + 1/(2n+1), which is not an integer, maybe positioning it to span n + 1 units in some direction?Wait, perhaps the key is that the side length is just enough to almost reach n + 1, but not quite. So, if we position the square such that it starts at 0 and goes to n + 1/(2n+1), it doesn't reach n + 1. However, if we rotate the square, we can make the projection along one axis longer, thereby covering n + 1 units in that direction, which would allow covering an additional column or row.Wait, for example, if the square is rotated so that its projection along the x-axis is n + 1 units, then even though the actual side length is less than n + 1, the projection is n + 1. How is that possible?Wait, the projection of the square onto the x-axis is equal to the side length times cos(theta) + side length times sin(theta), where theta is the angle of rotation. Wait, actually, the width of the rotated square is s*|cos(theta)| + s*|sin(theta)|. So if we set theta such that s*(cos(theta) + sin(theta)) = n + 1, then the projection would be n + 1. But s = n + 1/(2n+1). So:(n + 1/(2n+1))*(cos(theta) + sin(theta)) = n + 1We can solve for theta:cos(theta) + sin(theta) = (n + 1)/(n + 1/(2n+1)) = (n + 1)/( (n*(2n+1) + 1)/(2n+1) )) = (n + 1)*(2n + 1)/(2n^2 + n + 1)Hmm, simplifying the denominator: 2n^2 + n + 1. Let's compute this expression:(n + 1)*(2n + 1) = 2n^2 + 3n + 1Therefore, cos(theta) + sin(theta) = (2n^2 + 3n + 1)/(2n^2 + n + 1)This ratio is (2n^2 + 3n + 1)/(2n^2 + n + 1) = 1 + (2n)/(2n^2 + n + 1). Since n is positive, this is greater than 1 but less than, say, 1 + 1/n for large n.However, the maximum possible value of cos(theta) + sin(theta) is sqrt(2) ≈1.4142. So, if the ratio (2n^2 + 3n + 1)/(2n^2 + n + 1) is less than sqrt(2), then such a theta exists. Let's check for n=1:(2 + 3 +1)/(2 +1 +1) = 6/4=1.5 < sqrt(2)≈1.4142? Wait, 1.5>1.4142. So for n=1, this ratio is 1.5, which is greater than sqrt(2). Therefore, no solution for theta exists. For n=2:(2*4 + 6 +1)/(2*4 + 2 +1) = (8 +6 +1)/(8 +2 +1)=15/11≈1.3636 < sqrt(2). So here, it's possible. So for n >=2, maybe possible? Wait, check n=3:(2*9 +9 +1)/(2*9 +3 +1)= (18 +9 +1)/(18 +3 +1)=28/22≈1.2727 < sqrt(2). So yes, for n>=2, the ratio is less than sqrt(2), so there exists a theta such that the projection is n +1. For n=1, it's 1.5, which is greater than sqrt(2), so impossible. Therefore, for n >=2, rotating the square such that its projection on the x-axis is n +1 is possible. Therefore, such a square would span n +1 units in the x-direction, thereby covering n +2 lattice points in the x-direction (from 0 to n +1). Similarly, if we do the same for the y-direction, it would cover n +2 points. But wait, actually, the projection on the x-axis is n +1, but the actual coordinates covered depend on the position of the square.Wait, if the projection on the x-axis is n +1, but the square is rotated, the actual x-coordinates covered would be over an interval of length n +1. Therefore, if positioned from x=0 to x=n +1, it would cover n +2 lattice points in the x-direction. Similarly for y. But since the square is rotated, the overlapping in x and y might not be independent. So, the total number of lattice points covered might be (n +2)^2. But this seems too optimistic.But wait, the actual area of the square is (n +1/(2n+1))^2, which is less than (n +1)^2. Therefore, having (n +2)^2 lattice points would require the area to be at least that, but the actual area is smaller. Therefore, this is impossible. So, there must be a mistake in this reasoning.Wait, the key is that when you rotate the square, you can make the projection longer, but the actual number of lattice points covered is not necessarily the product of the projections. Because the square is rotated, some lattice points that lie in the projection might not be inside the square.Therefore, even though the projection on the x-axis is n +1, the actual number of x-coordinates covered by the square is n +1, but the y-coordinate for each x might vary. Therefore, it's not straightforward to compute the total number of lattice points as the product.Alternatively, perhaps using the concept of the covering number. The minimum number of lattice points that can cover a square of a given size. But here, we need the maximum.Alternatively, think of the square as a translated and rotated grid. Maybe using some lattice theory.Wait, maybe an example with n=2 will help. For n=2, the side length is 2 + 1/5 = 2.2. If I rotate this square such that its projection on the x-axis is 3 units (n +1=3). Then the angle theta satisfies 2.2*(cos(theta) + sin(theta))=3. So, cos(theta) + sin(theta)=3/2.2≈1.3636, which is possible because sqrt(2)≈1.4142>1.3636. So theta≈acos(1.3636 - sin(theta)). Wait, maybe compute theta numerically.Let me denote c = cos(theta), s = sin(theta). Then c + s = 3/2.2 ≈1.3636. Also, c^2 + s^2 =1. So, we have:c + s =1.3636c^2 + s^2 =1Let me solve for c and s. Let's set c + s = k, so k=1.3636. Then (c + s)^2 = c^2 + 2cs + s^2 =k^2. Therefore, 1 + 2cs =k^2. Therefore, cs=(k^2 -1)/2≈(1.3636^2 -1)/2≈(1.859 -1)/2≈0.859/2≈0.4295.Therefore, we have c and s such that c + s=1.3636 and c*s=0.4295. The solutions to the quadratic equation x^2 -1.3636x +0.4295=0. The discriminant is 1.3636^2 -4*0.4295≈1.859 -1.718≈0.141. Therefore, sqrt(0.141)=0.3755. Therefore, c=(1.3636 ±0.3755)/2. Therefore, c≈(1.3636 +0.3755)/2≈1.7391/2≈0.8695 or c≈(1.3636 -0.3755)/2≈0.9881/2≈0.494. Therefore, cos(theta)=0.8695 or 0.494. Then theta≈acos(0.8695)≈29.9 degrees or theta≈acos(0.494)≈60.4 degrees. Wait, but sin(theta) would then be sqrt(1 -c^2). Let's check for c=0.8695: sin(theta)=sqrt(1 -0.756)=sqrt(0.244)=0.494, so theta≈29.9 degrees. For c=0.494: sin(theta)=sqrt(1 -0.244)=sqrt(0.756)=0.8695, so theta≈60.4 degrees.Therefore, rotating the square by either 29.9 degrees or 60.4 degrees would result in the projection on the x-axis being 3 units. Now, if we position the square such that its projection on the x-axis spans from 0 to 3, then it would cover x=0,1,2,3. Similarly, if the projection on the y-axis also spans 3 units, then y=0,1,2,3. However, the actual square is rotated, so not all of these points will be inside the square.But perhaps some of them are. How can we estimate how many?Alternatively, the area of the square is 2.2^2=4.84. The number of lattice points inside can't exceed roughly the area plus some perimeter term. If we have 16 lattice points in the bounding box, but the area is only 4.84, then according to Pick's theorem, if the square is a lattice polygon, which it isn't necessarily, but approximately, the number of lattice points would be around Area + Perimeter/2 +1. Wait, but Pick's theorem requires the polygon to have vertices at lattice points. Since our square is rotated and scaled, its vertices are not at lattice points, so Pick's theorem doesn't apply directly.But as an approximation, the number of lattice points is roughly Area + O(Perimeter). For our square, the perimeter is 4*2.2=8.8. So the number of lattice points would be approximately 4.84 + 8.8/2 +1 ≈4.84 +4.4 +1≈10.24. So about 10 or 11 points. But the axis-aligned square would have (2 +1)^2=9 points. So maybe by rotating, we can cover 10 or 11 points? For n=2, perhaps the maximum is 10 or 11.But how to verify this?Alternatively, actually try to construct such a square. Let's take n=2, side length 2.2, rotated at 30 degrees (approximately). Place the center of the square at (1.5,1.5), so that it's between lattice points. Then, the square would extend from 1.5 - 1.1*cos(30) to 1.5 + 1.1*cos(30) in the x-direction, and similarly in the y-direction. cos(30)=sqrt(3)/2≈0.866, so 1.1*0.866≈0.9526. Therefore, the x-extent is from 1.5 -0.9526≈0.5474 to 1.5 +0.9526≈2.4526. Therefore, the x-coordinates covered are 1 and 2 (since 0.5474 <1 and 2.4526 <3). Similarly for y. Therefore, the lattice points would be (1,1), (1,2), (2,1), (2,2), which is 4 points. But that's way less than the axis-aligned case. Wait, that can't be right. Maybe positioning is important.Wait, perhaps if the square is placed so that its edges pass through lattice points. For example, if a square is rotated so that its sides pass through several lattice points, thereby increasing the count.Alternatively, consider a square that is positioned such that each side passes through a lattice point. For example, if we have a square with side length 2.2 rotated such that its sides pass through (0,0), (2,2), etc. But I need to make this precise.Alternatively, use the concept of the minimal number of lattice points on the boundary, but we want the maximal.Alternatively, think of the square as covering certain rows and columns. For an axis-aligned square, it's straightforward: from column 0 to column n, and row 0 to row n. But a rotated square might intersect more rows and columns. For example, a square rotated at 45 degrees with side length s will intersect approximately s*sqrt(2)/2 rows and columns. For s=2.2, this is about 1.555, so approximately 2 rows and columns. But that's similar to the axis-aligned case.Wait, perhaps there's a different approach. The problem gives a side length of n + 1/(2n+1). This seems specifically designed. Let me write the side length as n + ε, where ε=1/(2n+1). Notice that 2n+1 is the denominator, so 1/(2n+1) is a fraction that, when multiplied by 2n+1, gives 1. This might be a hint towards some periodic arrangement or something related to the grid.Alternatively, note that ε=1/(2n+1) is chosen such that n + ε = n +1/(2n+1). If we consider a square positioned such that it starts at (0,0) and extends to (n + ε, n + ε), then along each axis, it spans n + ε units. But if we instead place the square such that it is centered around a lattice point, then perhaps the sides will be positioned at ±(n + ε)/2 from the center. Therefore, the distance from the center to each side is (n + ε)/2. If the center is at a lattice point, then the sides are offset by (n + ε)/2. If (n + ε)/2 is not a half-integer, then the sides may not align with any grid lines.But since ε=1/(2n+1), then (n + ε)/2 = (n +1/(2n+1))/2 = (n(2n+1) +1)/(2(2n+1)) )= (2n^2 +n +1)/(2(2n+1)). Not sure if that helps.Wait, perhaps the key is that 2n+1 is the denominator in ε. Let me consider that the side length is n +1/(2n+1). If we look at the continued fraction or some other property.Alternatively, think of the Farey sequence or rational approximations. Maybe the fraction 1/(2n+1) is chosen to align with some grid lines when the square is rotated.Alternatively, note that (2n+1)ε=1. So if we have a square with side length n + ε, then (2n+1)(n + ε) = (2n+1)n +1=2n^2 +n +1. This might be a hint towards some tiling or periodicity.Alternatively, consider that the square can be partitioned into (2n+1)x(2n+1) smaller squares each of side length 1/(2n+1). But since the side length is n +1/(2n+1), that equals (n(2n+1) +1)/(2n+1) = (2n^2 +n +1)/(2n+1). So, the square's side can be divided into 2n^2 +n +1 segments each of length 1/(2n+1). But I don't see how this helps.Wait, maybe the number of lattice points covered by the square is related to the number of unit squares it intersects. There's a theorem that says the number of unit squares a line segment between two lattice points intersects is related to the gcd of the differences in coordinates. But again, not sure.Alternatively, think of the square as a region in the plane and count the number of lattice points inside using integration or something. But that seems too vague.Wait, going back to the original problem: the square can be placed anywhere (translated and rotated) to cover as many lattice points as possible. The question is, what's the maximum number?For an axis-aligned square of side length n + ε, where ε=1/(2n+1), as we saw earlier, it can cover (n +1)^2 lattice points. If we rotate the square, can we cover more?Alternatively, maybe yes. If we rotate the square such that its corners are at lattice points, then the square might cover more points. For instance, if the square is rotated 45 degrees and scaled so that its corners are at integer coordinates. But our square has a specific side length, so scaling is not an option.Alternatively, use the fact that the maximum number of lattice points in a square is related to the determinant of the lattice basis, but I'm not sure.Alternatively, look for existing theorems or formulas. I recall that the maximum number of lattice points in a square of side length s is roughly (s)^2, but adjusted by the orientation. However, since we're allowed any orientation, perhaps the maximum is achieved when the square is aligned with the grid. But I need to confirm.Wait, here's a different thought. The side length n +1/(2n+1) is just slightly larger than n. So, an axis-aligned square of this side length can be placed from (0,0) to (n +1/(2n+1), n +1/(2n+1)). The integer points in this square are all (i,j) where 0 <=i,j <=n. Because n +1/(2n+1) <n +1, right? Since 1/(2n+1) <1. So, the maximum integer coordinate in each direction is n. Therefore, the number of lattice points is (n +1)^2. If we rotate the square, maybe we can have some points where i or j exceed n, but given the side length is just over n, but rotated, maybe the square can reach n +1 in some coordinate? For example, if we rotate the square so that one corner is at (n +1/(2n+1),0) and the opposite corner is at (0,n +1/(2n+1)), then the diagonal of the square would be along the line y = -x + n +1/(2n+1). The length of the diagonal is s*sqrt(2), where s is the side length. So, s*sqrt(2) = (n +1/(2n+1))*sqrt(2). The distance between (n +1/(2n+1),0) and (0,n +1/(2n+1)) is sqrt( (n +1/(2n+1))^2 + (n +1/(2n+1))^2 ) = (n +1/(2n+1))sqrt(2). So, this matches the diagonal length. Therefore, such a rotated square would have its corners at (n +1/(2n+1),0), (0,n +1/(2n+1)), (-n -1/(2n+1),0), and (0,-n -1/(2n+1)). But this square is centered at the origin, but its corners extend beyond n in both x and y directions. However, the side length is n +1/(2n+1), so each side is length s =n +1/(2n+1). However, the distance from the center to each corner is s*sqrt(2)/2. But in this case, the corners are at a distance of n +1/(2n+1) from the center along the axes. Wait, maybe I confused the diagonal with the side length.Wait, if the square is rotated so that its diagonal is along the x-axis, then the side length is s, and the diagonal length is s*sqrt(2). But in the previous example, the diagonal was along the line y=-x, with length (n +1/(2n+1))sqrt(2). Therefore, the side length would be n +1/(2n+1). Therefore, the square with corners at (n +1/(2n+1),0), etc., has side length n +1/(2n+1). Wait, but the distance between (n + ε, 0) and (0, n + ε) is sqrt( (n + ε)^2 + (n + ε)^2 )= (n + ε)sqrt(2), which is the diagonal length. Therefore, the side length s must satisfy s*sqrt(2)= (n + ε)sqrt(2), so s= n + ε. Therefore, that square has side length n + ε, rotated 45 degrees. So, in this case, the square is rotated 45 degrees, with its edges going from (n + ε,0) to (0, n + ε), etc.But how many lattice points does this square cover? The square is diamond-shaped, centered at the origin, with vertices at (n + ε,0), (0,n + ε), (-n -ε,0), and (0,-n -ε). The edges are along the lines y= -x + c, where c= n + ε, etc. To count the lattice points inside this square, we can note that the square is the set of all (x,y) such that |x| + |y| <= n + ε. But since ε=1/(2n+1), which is a fraction, the inequality is |x| + |y| <= n + 1/(2n+1). The number of lattice points (i,j) with |i| + |j| <= n + 1/(2n+1). Since |i| and |j| are integers, |i| + |j| is an integer or half-integer depending on n. Wait, |i| + |j| must be an integer because |i| and |j| are integers. Therefore, |i| + |j| <= n + 1/(2n+1) implies |i| + |j| <= n, because n + 1/(2n+1) is not an integer. Therefore, the number of lattice points is the same as the number of points with |i| + |j| <=n. Which is a diamond shape (a square rotated 45 degrees) with layers. The number of points is (2n +1)^2. Wait, no. Wait, for each k from 0 to n, the number of points with |i| + |j|=k is 4k (for k >0), but actually, when considering all quadrants, it's 1 + 4*sum_{k=1}^n k =1 +4*(n(n +1))/2=1 + 2n(n +1). But in this case, since we are considering |i| + |j| <=n, it's the sum from k=0 to n of (4k +1) when k>0, but actually, the number of lattice points is (2n +1)^2. Wait, no. Wait, let's compute for n=1: points with |i| + |j| <=1. These are (0,0), (1,0), (-1,0), (0,1), (0,-1). Total 5 points. For n=2: (0,0), (±1,0), (0,±1), (±2,0), (0,±2), (±1,±1). Total 1 +4 +4 +4=13. Wait, this seems like (n +1)^2 +n^2. For n=1, 5= (1+1)^2 +1^2=4 +1=5. For n=2, 13= (2+1)^2 +2^2=9 +4=13. So, in general, the number of lattice points with |i| + |j| <=n is (n +1)^2 +n^2=2n^2 +2n +1. Wait, for n=1, 2*1 +2*1 +1=5, which matches. For n=2, 2*4 +4 +1=13, yes. So the formula is 2n^2 +2n +1. But this is the number of lattice points inside a diamond (rotated square) with side length sqrt(2)*n, but our square has side length n +1/(2n+1). However, in our case, the rotated square has side length n + ε, which is larger than n but less than n +1. Therefore, the diamond |i| + |j| <=n + ε would include more points than |i| + |j| <=n. However, since ε=1/(2n+1), which is 1 over an odd integer, and |i| + |j| must be integer, then |i| + |j| <=n + ε is equivalent to |i| + |j| <=n, because n + ε is not an integer. Therefore, the number of lattice points in the rotated square is still 2n^2 +2n +1. However, the axis-aligned square covers (n +1)^2 lattice points. Comparing the two:For n=1: rotated square has 5 points, axis-aligned has 4. Wait, so rotated covers more. For n=2: rotated has 13, axis-aligned has 9. So rotated square covers more points. Therefore, in this case, rotating the square allows covering more lattice points. But this seems to contradict my earlier thought that the axis-aligned square has (n +1)^2 points. But in reality, when rotated, the diamond-shaped square covers more points.Wait, but this rotated square is different from the square in the problem. The problem states a square of side length n + ε, not a diamond with |i| + |j| <=n + ε. Wait, in the rotated square I described earlier, the side length is n + ε, but the number of points inside is 2n^2 +2n +1. However, the area of this rotated square is (n + ε)^2. For example, when n=1, area=(1 +1/3)^2≈1.777, and the number of lattice points is 5. According to Pick's theorem, if it were a lattice polygon, Area =I + B/2 -1. But since it's not a lattice polygon, we can't apply it directly. However, the count of 5 points with area≈1.777 is possible because the points are mostly on the boundary.But this suggests that for the rotated square, even though the area is smaller than the axis-aligned case, it can cover more lattice points. Therefore, the maximum number of lattice points might indeed be higher when the square is rotated.Wait, but in the example for n=1, the axis-aligned square of side length 1 +1/3≈1.333 covers 4 points, while the rotated square covers 5 points. Therefore, rotating gives more points. Similarly, for n=2, the axis-aligned covers 9, rotated covers 13. Therefore, the rotated square seems better.But wait, the rotated square I'm considering is a diamond with vertices at (n + ε,0), etc., which has side length n + ε, but the distance from the center to each vertex is n + ε. Wait, no. If the side length is s, then the distance from the center to each vertex is s*sqrt(2)/2. But in our case, the distance from the center (0,0) to (n + ε,0) is n + ε, so that would mean the side length is (n + ε)*sqrt(2). But we were told the side length is n + ε. Therefore, my previous assumption is incorrect. The square with vertices at (n + ε,0), (0,n + ε), etc., has side length (n + ε)*sqrt(2), not n + ε. Therefore, this is a different square.Therefore, my mistake was in conflating the side length with the distance from the center. The actual square with side length n + ε, rotated 45 degrees, would have vertices closer to the center. Let me correct this.If the square has side length s= n + ε, and it's rotated 45 degrees, then the distance from the center to each vertex is s*sqrt(2)/2. Therefore, the vertices would be at (s*sqrt(2)/2,0), etc. But to have these vertices at integer coordinates, s*sqrt(2)/2 must be integer, which is not our case. However, we can still compute the number of lattice points inside such a square.Alternatively, perhaps the maximum number of lattice points is indeed achieved by the rotated square, but I need to verify the count.Wait, let's take n=1 again. Side length s=1 +1/3≈1.333. If we rotate this square by 45 degrees, its bounding box would have side length s*sqrt(2)≈1.885. So, the bounding box would span from approximately -0.9425 to +0.9425 in x and y, covering x=-0.9425 to 0.9425, which would include x=-0.9425 to 0.9425, which is less than 1 unit in each direction. Therefore, the lattice points covered would be (0,0) only. Wait, that can't be right. No, the center of the square is at (0,0), rotated 45 degrees, with side length 1.333. The actual corners are at (±s/2, ±s/2) rotated 45 degrees, which would be (±0.6665, ±0.6665) rotated by 45 degrees. Wait, no, the rotation of a square centered at the origin with side length s would have vertices at (s/2, s/2), (-s/2, s/2), etc., rotated by 45 degrees. Wait, maybe my coordinate system is messed up.Actually, to rotate a square centered at the origin with side length s, the vertices before rotation would be at (s/2, s/2), (-s/2, s/2), etc. Rotating these points by 45 degrees would give new vertices. But calculating the exact positions requires some computation.Alternatively, forget the rotation. Let's think of the square as diamond-shaped, defined by |x| + |y| <= s/2*sqrt(2). But this is getting too vague.Perhaps it's better to refer back to the original examples. For n=1, the axis-aligned square covers 4 points, but the rotated square (if possible) covers 5. Similarly, for n=2, axis-aligned covers 9, rotated covers 13. If this pattern continues, the number of points for the rotated square is 2n^2 +2n +1. Therefore, the maximum number is 2n^2 +2n +1. But wait, for n=1, 2*1 +2*1 +1=5, which matches. For n=2, 2*4 +4 +1=13, as before. But this seems like a quadratic function, 2n^2 +2n +1. However, we need to verify whether such a rotated square with side length n +1/(2n+1) can indeed cover that many points.Wait, for the diamond |x| + |y| <=k, the number of lattice points is indeed 2k^2 +2k +1 when k is integer. But in our case, k= n + ε, which is not integer. Therefore, the number of lattice points would still be 2n^2 +2n +1, since |x| + |y| <=n + ε <n +1, and |x| + |y| must be integer. Therefore, the number of lattice points inside is the same as for k=n. Therefore, the rotated square doesn't gain any points. Therefore, this approach doesn't work.But earlier, for n=1, we saw that a rotated square can cover 5 points versus axis-aligned 4. How is that possible?Wait, perhaps when the square is rotated, it can actually include lattice points that are outside the axis-aligned bounding box. For example, take n=1, side length≈1.333. If we rotate the square by 45 degrees and center it at (0.5,0.5), then the square might cover more points. Let's compute.The square has side length≈1.333. When rotated 45 degrees, the half-diagonal is (1.333)/sqrt(2)≈0.9428. So, the square extends from 0.5 -0.9428 to0.5 +0.9428 in both x and y. Approximately, x from≈-0.4428 to≈1.4428. Similarly for y. Therefore, the integer points covered are x=0,1 and y=0,1. So (0,0), (0,1), (1,0), (1,1). That's 4 points. So no gain.But wait, if we position the square differently. Suppose we place one corner at (0,0) and the square extends into the positive quadrant. The square is rotated so that its sides are along the lines y= x and y= -x + s. For s=1.333. Then, the square would have vertices at (0,0), (s,0), (s,s), and (0,s). Wait, but rotated?Wait, no. If we have a square with one corner at (0,0) and side length s, rotated by 45 degrees, then the coordinates get more complicated. Alternatively, place the square such that its sides are not aligned with the axes but still in the positive quadrant.Alternatively, consider the square with vertices at (0,0), (s*cos(theta), s*sin(theta)), (-s*sin(theta), s*cos(theta)), and (s*cos(theta) -s*sin(theta), s*sin(theta) +s*cos(theta)). This is a rotated square. But calculating which lattice points are inside would be tedious.Alternatively, think of the square as covering certain rows and columns. For a rotated square, each row and column can be intersected twice, leading to more lattice points. But I need to formalize this.Wait, another idea. The maximum number of lattice points in a square is related to how many grid lines it intersects. For each vertical grid line x =k, the square can intersect it twice, leading to a segment inside the square. The number of lattice points on this segment would be the number of integer y's such that (k,y) is inside the square. Similarly for horizontal lines.But to maximize the total number of lattice points, we need to maximize the number of vertical and horizontal grid lines intersected by the square, and within each, the number of y's or x's.For an axis-aligned square, it intersects n +2 vertical lines (from 0 to n +1, but the square of side length n + ε doesn't reach n +1, so only n +1 vertical lines. Each vertical line has n +1 points inside. So total points (n +1)^2.For a rotated square, the number of vertical lines intersected can be more. For example, a square rotated at a slight angle will intersect more vertical and horizontal lines. Each intersection with a vertical line can contribute up to some number of y's, depending on the height of the intersection.But calculating this precisely is complex. However, there's a theorem by van Wamelen that gives the maximum number of lattice points in a square of given side length. But I don't recall the exact statement.Alternatively, recall that the maximum number of lattice points in a square with side length s is at most O(s^2), which is trivial, but we need the exact maximum.Alternatively, consider that the problem gives a side length of n +1/(2n+1). This seems tailored such that when the square is rotated appropriately, it can cover an additional "layer" of lattice points compared to the axis-aligned case.For example, consider that the side length n +1/(2n+1) is just enough so that when the square is rotated by a certain angle, it can cover n +1 rows and columns in both directions, thereby achieving (n +2)^2 lattice points. However, as we saw earlier, the area is less than (n +1)^2, so this is impossible. Therefore, the next idea is that rotating the square can allow covering more lattice points along the boundary, thereby increasing the total count beyond (n +1)^2.Alternatively, using the concept that a rotated square can have more intersections with the grid lines, thereby creating more opportunities for lattice points to be inside. For example, each side of the square can intersect multiple grid lines, and if the intersections are at lattice points, then those points are included.However, positioning the square such that its sides pass through lattice points can maximize the boundary points, thereby increasing the total count via Pick's theorem.Recall that Pick's theorem states A = I + B/2 -1, where A is area, I is interior lattice points, and B is boundary lattice points. So, total lattice points is I + B = A + B/2 +1. Therefore, to maximize I + B, we need to maximize B.Thus, a square with a high number of boundary lattice points will have a higher total lattice points. Therefore, if we can find a square with side length n +1/(2n+1) that has a large B, then the total points will be higher.To maximize B, the number of boundary lattice points, the square should have edges that pass through many lattice points. Each edge can pass through gcd(delta_x, delta_y) +1 lattice points, where delta_x and delta_y are the differences in coordinates along the edge.Therefore, to maximize B, we want each edge to pass through as many lattice points as possible. For that, the direction of the edges should be such that gcd(delta_x, delta_y) is maximized.Suppose we choose the direction of the edges to be (k, 1), where k is an integer. Then gcd(k,1)=1, so each edge would pass through 2 lattice points (the endpoints). If we choose a direction with a higher gcd, say (2,2), then gcd(2,2)=2, so the edge would pass through 3 lattice points. But the side length must be n +1/(2n+1).Alternatively, if the side vector is (m, m), then the length is m*sqrt(2). If we set m*sqrt(2) =n +1/(2n+1), then m=(n +1/(2n+1))/sqrt(2). But m needs to be rational to have lattice points along the edge. However, this seems difficult because of the sqrt(2).Alternatively, find integers a and b such that the vector (a,b) has length approximately n +1/(2n+1), and gcd(a,b)=d is maximized. The number of lattice points on the edge would be d +1.But this approach is too vague without concrete calculations.Wait, going back to the problem's side length: n +1/(2n+1). Let's denote s =n +1/(2n+1). Let's compute s for small n:For n=1: s=1 +1/3=4/3≈1.333For n=2: s=2 +1/5=11/5=2.2For n=3:3 +1/7≈3.1429Etc.Notice that for n=1, s=4/3. If we create a square with side length 4/3, rotated such that its edges are along the vectors (1,1), then the length of (1,1) is sqrt(2), which is≈1.414, which is larger than 4/3≈1.333. So that doesn't fit. Alternatively, find a vector with smaller length.Alternatively, take the vector (3,1). Its length is sqrt(10)≈3.162. For n=3, s≈3.1429, which is close. But not sure.Alternatively, use continued fractions to find a rational approximation of s with a small denominator.But this might not be the right path.Alternatively, think of the square as covering a certain number of rows and columns. For example, an axis-aligned square covers n +1 rows and columns. A rotated square might cover more rows and columns, but each row and column would have fewer points inside.But the total number might be similar or higher. For example, if a rotated square covers n +2 rows and n +2 columns, but each row has n +1 points inside, then total is (n +2)(n +1). But this is larger than (n +1)^2.But this is speculation. Let me try n=1 again.For n=1, s=4/3≈1.333. If we rotate the square by some angle, can we cover more than 4 lattice points?Suppose we rotate the square so that its sides are not aligned with the grid. Place the square such that its bottom-left corner is at (0.5, 0.5), and it extends to (0.5 + s*cos(theta), 0.5 + s*sin(theta)). Choosing theta such that the square covers more lattice points.But without concrete coordinates, it's hard to tell. Alternatively, use a coordinate system where the square is centered at (0.5,0.5), sides of length 4/3, rotated by some angle. The four vertices would be at (0.5 + (4/3)/2*cos(theta) - (4/3)/2*sin(theta), 0.5 + (4/3)/2*sin(theta) + (4/3)/2*cos(theta)), etc. This is complicated.Alternatively, use a different approach. The problem specifies the side length is n +1/(2n+1). Notice that 2n+1 is the denominator, which is an odd number. Maybe there's a way to position the square such that it touches lattice points at intervals of 1/(2n+1), thereby covering multiple points.Alternatively, notice that 1/(2n+1) is the smallest unit in a grid partitioned into (2n+1)x(2n+1) subdivisions. Therefore, if the square is positioned to align with this fine grid, it can cover additional points.But I'm not sure.Wait, another idea: the side length is n +1/(2n+1). If we multiply the entire plane by (2n+1), then the lattice points become ( (2n+1)Z )^2, and the square becomes a square of side length (2n+1)(n) +1. Which is an integer: 2n^2 +n +1. Therefore, scaling back, the original square of side length n +1/(2n+1) can be thought of as a square of integer side length 2n^2 +n +1 in the scaled grid. Therefore, in the scaled grid, the square can cover (2n^2 +n +2)^2 lattice points. But when scaling back down, each lattice point in the scaled grid corresponds to a point with coordinates multiples of 1/(2n+1). Therefore, the number of original lattice points covered would be related to the number of these fine grid points inside the square.But I'm not sure if this helps. This might be overcomplicating.Alternatively, consider that the square of side length n +1/(2n+1) can be partitioned into n^2 small squares each of side length 1, plus a strip of width 1/(2n+1). Therefore, the main part of the square covers n^2 unit squares, each contributing 4 lattice points, but overlapping. But this seems unproductive.Alternatively, think of the square as a slightly enlarged n x n square. The extra 1/(2n+1) allows it to cover an additional column and row. Wait, but the side length is n + ε, where ε=1/(2n+1). If we place the square so that it starts at x=0 and goes to x=n + ε, then in the x-direction, it covers n +1 columns (0 to n), since ε <1. Similarly for y. So total lattice points (n +1)^2.But if we could somehow shift the square so that the extra ε allows it to cover partially into the next column and row, but since ε is very small, it's unlikely. However, if the square is rotated, the projection on both axes increases, so maybe the square can span n +1 columns and rows even though its side length is just n + ε.Wait, the projection of the square onto the x-axis is s*cos(theta) + s*sin(theta). To make this projection equal to n +1, we set s*(cos(theta) + sin(theta))=n +1. As before, this is possible for n >=2. Therefore, the square can span n +1 columns and rows, but since it's rotated, each column and row may have multiple points inside.But the total number of lattice points would then be up to (n +1)^2, similar to the axis-aligned case. However, due to rotation, some points might be counted in both projections, leading to overlaps. Therefore, it's unclear if this provides an advantage.Alternatively, if the square is rotated such that its sides pass through lattice points every 1/(2n+1) units. For example, if the square is aligned with a grid that's finer by a factor of 2n+1, then each step of 1/(2n+1) corresponds to a lattice point in the fine grid. But this is too vague.Given that I'm going in circles here, maybe I should look for a pattern or formula.For the axis-aligned square, we get (n +1)^2 points.For the rotated square, as in the diamond example, we saw higher counts, but that required a different side length.But in our problem, the side length is specifically n +1/(2n+1). The key might be that this side length allows the square to be positioned such that it covers n +1 columns and rows when rotated, thereby achieving (n +1)^2 points, same as axis-aligned, but actually, no, the rotated square's count might differ.Wait, here's a different approach inspired by the side length. The side length is n +1/(2n+1). Notice that 2n+1 is an odd integer. If we tile the plane with a grid spaced 1/(2n+1) units apart, the lattice points would be multiples of 1/(2n+1). The side length of the square is n +1/(2n+1) = (n(2n+1) +1)/(2n+1) = (2n^2 +n +1)/(2n+1). This is an integer in the fine grid: 2n^2 +n +1. Therefore, in the fine grid, the square has integer side length and can be aligned with the grid, thereby covering (2n^2 +n +2)^2 fine lattice points. Translating back to the original lattice, each fine lattice point corresponds to a point (k/(2n+1), m/(2n+1)), which is a lattice point only if k and m are multiples of 2n+1. Therefore, the number of original lattice points covered by the square is the number of fine lattice points that are also original lattice points. This is equivalent to the number of fine lattice points with coordinates divisible by 2n+1, which is ((2n^2 +n +1)/ (2n+1) +1)^2 = (n +1 +1)^2 = (n +2)^2. But this is only if the square is axis-aligned in the fine grid.Wait, let's clarify:In the scaled grid where each unit is 1/(2n+1), the square has side length 2n^2 +n +1. Therefore, axis-aligned, it covers (2n^2 +n +1 +1)^2 = (2n^2 +n +2)^2 points in the fine grid. But each original lattice point corresponds to a point in the fine grid where both coordinates are multiples of 2n+1. Therefore, the number of original lattice points inside the square is the number of fine grid points that are multiples of 2n+1 in both coordinates. This is floor((2n^2 +n +1)/ (2n+1)) +1)^2.Compute floor((2n^2 +n +1)/(2n+1)):Divide 2n^2 +n +1 by 2n+1:2n^2 +n +1 = (2n+1)(n) -n +1 = 2n^2 +n -n +1 = 2n^2 +1. Wait, that's incorrect. Let's perform polynomial division.Divide 2n^2 +n +1 by 2n+1:2n^2 +n +1 ÷ 2n+1:First term: n. Multiply 2n+1 by n: 2n^2 +n. Subtract from dividend: (2n^2 +n +1) - (2n^2 +n) =1. Remainder:1. Therefore, 2n^2 +n +1 = (2n+1)*n +1. Therefore, floor((2n^2 +n +1)/(2n+1))=n. Therefore, the number of original lattice points is (n +1)^2. Which matches the axis-aligned case.Therefore, this approach doesn't give any new information.Given all these considerations and the time I've spent without reaching a concrete conclusion, I think the answer is likely (n +1)^2 +4n =n^2 +6n +1, but I need to verify.Wait, for n=1, this gives 1 +6 +1=8, which is more than the axis-aligned 4 and the rotated 5. So that's incorrect.Alternatively, maybe the formula is (n +1)^2 +4n. For n=1, 4 +4=8, which is too high. For n=2, 9 +8=17, which is higher than the rotated 13.Alternatively, recall that for the diamond-shaped square (rotated), the number of points is 2n^2 +2n +1. For n=1, 2+2+1=5, which works. For n=2, 8+4+1=13. So if this is the case, then perhaps the maximum number of lattice points is 2n^2 +2n +1.But how does this relate to the side length?The diamond-shaped square with |x| + |y| <=k has area 2k^2. Setting 2k^2=(n +1/(2n+1))^2. Solving for k: k=(n +1/(2n+1))/sqrt(2). But k must be such that |x| + |y| <=k includes certain lattice points. However, since k is not an integer, the number of lattice points is the same as floor(k) +1? Not sure.Alternatively, perhaps the diamond-shaped square with side length n +1/(2n+1) rotated 45 degrees can cover 2n^2 +2n +1 lattice points. For example, for n=1, it covers 5 points, which is more than the axis-aligned square. For n=2, 13 points. Therefore, this might be the maximum.But how to confirm that such a rotated square with side length n +1/(2n+1) can indeed cover these many points.Wait, for n=1, side length=4/3≈1.333. The diamond |x| + |y| <=k. The area of the diamond is 2k^2. Set 2k^2=(4/3)^2=16/9≈1.777. Then k^2=8/9≈0.888, so k≈0.9428. Therefore, the diamond |x| + |y| <=0.9428 includes only the origin (0,0), since any other point would have |x| + |y| >=1. Therefore, only 1 point. But we previously thought it could cover 5 points, which is incorrect. Therefore, my earlier reasoning was flawed.Therefore, the diamond-shaped square with side length n +ε doesn't cover 2n^2 +2n +1 points. Instead, it covers only the points with |x| + |y| <=n, since n +ε <n +1. Therefore, the number of points is 2n^2 +2n +1.Wait, but for n=1, this gives 5 points, but the diamond |x| + |y| <=1 has 5 points. However, the square with side length 4/3 rotated 45 degrees is smaller than the diamond |x| + |y| <=1, so it can't cover all 5 points. Therefore, there's a contradiction.This indicates that my initial assumption was wrong: rotating the square doesn't allow covering more points than the axis-aligned case. Therefore, the maximum number of lattice points is indeed (n +1)^2, achieved by the axis-aligned square.But wait, for n=1, the axis-aligned square covers (1 +1)^2=4 points, but we saw earlier that a rotated square of the same side length might cover 5 points. However, based on the area, the rotated square would have area≈1.777, which is larger than the axis-aligned square's area ( (4/3)^2≈1.777 ). Wait, both squares have the same area. Therefore, Pick's theorem would apply similarly. But a rotated square can indeed have more points if its boundary has more lattice points.But in the case of the diamond |x| + |y| <=1, which has area 2, the number of lattice points is 5. If we scale it down to area≈1.777, the number of lattice points would decrease. Therefore, my previous example was incorrect.Therefore, it seems that the axis-aligned square is the best configuration, giving (n +1)^2 lattice points.But why for n=1, the axis-aligned square of side length 4/3 covers 4 points, while the diamond |x| + |y| <=1 has area 2 and 5 points. But our square has area≈1.777, which is less than 2, so it cannot cover all 5 points. Therefore, the maximum number remains 4.Therefore, after all this back-and-forth, I think the correct answer is (n +1)^2, the same as the axis-aligned square. The rotation doesn't help because the side length is not sufficient to cover additional lattice points beyond the axis-aligned case.Therefore, the maximum number of lattice points is boxed{(n + 1)^2}.Final AnswerThe maximum number of lattice points is boxed{(n + 1)^2}.</think>Given a positive integer ( n ), we need to determine the maximum number of lattice points a square of side length ( n + frac{1}{2n+1} ) may cover.1. Axis-Aligned Square Consideration: - An axis-aligned square with side length ( n + frac{1}{2n+1} ) spans from ( 0 ) to ( n + frac{1}{2n+1} ) in both the x and y directions. - The side length ( n + frac{1}{2n+1} ) is slightly larger than ( n ) but less than ( n + 1 ). - This means the square will cover integer coordinates from ( 0 ) to ( n ) in both directions, resulting in ( (n + 1) ) points along each axis. - Therefore, the total number of lattice points covered by an axis-aligned square is ( (n + 1)^2 ).2. Rotated Square Consideration: - Rotating the square might increase the number of lattice points covered if the projection on the axes spans more columns and rows. - However, the side length ( n + frac{1}{2n+1} ) is not sufficient to span ( n + 1 ) units in any direction when rotated, as the projection would still be less than ( n + 1 ). - The area of the square is ( (n + frac{1}{2n+1})^2 ), which is approximately ( n^2 + 1 ). According to Pick's theorem, the number of lattice points is related to the area and perimeter, but the exact maximum is achieved by the axis-aligned configuration.3. Conclusion: - The axis-aligned square covers ( (n + 1)^2 ) lattice points]

🔑:1. Lemma: Among any 5 lattice points, there are at least two of them placed at a distance of at least sqrt{5}. This can be shown by considering the Euclidean distance between any two points in a Cartesian plane. If we assume a counterexample, we can use the Euclidean distance formula to show that the sets representing the possible placements of the points intersect to form an empty set, which is a contradiction.2. Observation: Notice that frac{1}{3} < -1 + sqrt{frac{5}{2}}, which implies that sqrt{5} > (1 + frac{1}{3})sqrt{2}. The right-hand side of this inequality represents the diagonal of a square with side length 1 + frac{1}{3}. If this square were to cover 5 lattice points, the distance of its diagonal would be at least the maximum distance any of these points may have, which would contradict the lemma. Therefore, the number of lattice points that a square of side 1 + frac{1}{3} may cover is at most 4. This is our first key result.3. General Case for n ge 2: Let n be a natural number such that n ge 2, and let S be a square of side length n + frac{1}{2n+1} that contains exactly i lattice points in its interior and b lattice points on its boundary. Let H be the convex hull of the lattice points covered by S. Since H is bounded by S, the area of H is at most the area of S.4. Pick's Theorem Application: By Pick's theorem, we have: [ i + b = mathcal{A}_H + frac{b}{2} + 1 ] Given that the area of H is at most the area of S, and the perimeter of H is at most the perimeter of S, we can write: [ i + b le leftlfloor left( n + frac{1}{2n+1} right)^2 + frac{1}{2} leftlfloor 4 left( n + frac{1}{2n+1} right) rightrfloor + 1 rightrfloor ]5. Simplifying the Expression: [ left( n + frac{1}{2n+1} right)^2 = n^2 + frac{2n}{2n+1} + frac{1}{(2n+1)^2} ] [ frac{1}{2} leftlfloor 4 left( n + frac{1}{2n+1} right) rightrfloor = 2n + 1 ] Therefore: [ i + b le leftlfloor n^2 + frac{2n}{2n+1} + frac{1}{(2n+1)^2} + 2n + 1 rightrfloor ] [ le leftlfloor (n+1)^2 + frac{1 + 4n}{1 + 4n + 4n^2} rightrfloor = (n+1)^2 ]6. Verification: Consider the sequence of squares characterized by the Cartesian vertices: (0,0), (0, n + frac{1}{2n+1}), (n + frac{1}{2n+1}, n + frac{1}{2n+1}), (n + frac{1}{2n+1}, 0) for any natural number n. These indeed cover (n+1)^2 lattice points.With the above steps, we have shown that the number of lattice points that a square of side n + frac{1}{2n+1} may cover is at most (n+1)^2.The final answer is boxed{(n+1)^2}

🧠:[<think>Okay, so I need to figure out how many common tangent lines there are between two circles, C₁ and C₂. The answer choices are 1, 2, 3, or 4 lines. Hmm, let me recall what determines the number of common tangents between two circles. I think it has to do with their positions relative to each other—like if they're separate, overlapping, one inside the other, etc. First, maybe I should find out the centers and radii of both circles. The equations given are in general form, so I need to rewrite them in standard form by completing the squares.Starting with circle C₁: x² + y² + 2x - 6y = 0.Let me group the x terms and y terms:(x² + 2x) + (y² - 6y) = 0.Completing the square for x: Take the coefficient of x, which is 2, divide by 2 to get 1, square it to get 1. So add and subtract 1:(x² + 2x + 1 - 1) + (y² - 6y) = 0.Similarly for y: Coefficient of y is -6. Divide by 2: -3, square it: 9. Add and subtract 9:(x² + 2x + 1 - 1) + (y² - 6y + 9 - 9) = 0.Simplify:(x + 1)² - 1 + (y - 3)² - 9 = 0 → (x + 1)² + (y - 3)² = 10.So the center of C₁ is (-1, 3) and the radius is √10.Now circle C₂: x² + y² -4x + 2y + 4 = 0.Group x and y terms:(x² -4x) + (y² + 2y) = -4.Completing the square for x: Coefficient of x is -4. Divide by 2: -2, square it: 4. Add and subtract 4:(x² -4x +4 -4) + (y² +2y) = -4.For y: Coefficient is 2. Divide by 2: 1, square it: 1. Add and subtract 1:(x² -4x +4 -4) + (y² +2y +1 -1) = -4.Simplify:(x - 2)² -4 + (y +1)² -1 = -4 → (x - 2)² + (y +1)² -5 = -4 → (x - 2)² + (y +1)² = 1.Therefore, center of C₂ is (2, -1) and radius is √1 = 1.Now, let's find the distance between the centers. The centers are (-1, 3) and (2, -1). Using the distance formula:Distance = √[(2 - (-1))² + (-1 - 3)²] = √[(3)² + (-4)²] = √[9 + 16] = √25 = 5.So the distance between centers is 5 units. Now, the radii are √10 ≈ 3.16 and 1. Let's see the different cases for common tangents:1. If one circle is entirely inside another and they don't touch, there are 0 common tangents.2. If one circle is inside another and they touch internally, there's 1 common tangent.3. If the circles intersect at two points, there are 2 common tangents.4. If the circles are separate (neither overlapping nor touching), there are 4 common tangents: 2 direct and 2 transverse.5. If they touch externally, there are 3 common tangents (the point of contact plus two others).Wait, but let me verify. When two circles are externally tangent, they have three common tangents: two external ones and one at the point of contact. When they are separate, four tangents. When intersecting, two. When internally tangent, one. When one inside the other without touching, none. So in our case, let's check the sum and difference of radii compared to the distance between centers.Sum of radii: √10 + 1 ≈ 3.16 + 1 = 4.16. Distance between centers: 5.Since the distance (5) is greater than the sum of the radii (~4.16), the circles are separate. Therefore, there should be 4 common tangents. So the answer is D: 4 lines. Wait, but let me check again.Alternatively, maybe I made a mistake in the radii or the distance. Let's confirm:Radius of C₁: √10 ≈ 3.16. Radius of C₂: 1. Sum: ~4.16. Distance between centers: 5, which is more than the sum. So yes, they are separate circles. Therefore, 4 common tangents. So answer D.But wait, the options are A:1, B:2, C:3, D:4. So D is 4 lines. But let me make sure. Wait, sometimes there's a case where even if the circles are separate, if they are not too far apart, maybe the number of tangents is different? Wait no, the general rule is:- If circles are separate (distance > sum of radii): 4 common tangents.- If they are externally tangent (distance = sum): 3.- If intersecting (distance between sum and difference): 2.- If internally tangent (distance = difference): 1.- If one inside the other (distance < difference): 0.So in this case, distance is 5, sum is √10 +1 ≈4.16. So 5>4.16, so separate circles. Therefore, 4 common tangents. So answer D. But wait, let me compute √10 +1 exactly. √10 is approximately 3.1623, so 3.1623 +1≈4.1623. 5 is definitely larger. So yes, they are separate. Hence 4 tangents. So the answer is D. But wait, maybe I made a mistake in the calculation. Let me check again.Wait, centers are (-1,3) and (2,-1). Difference in x: 2 - (-1)=3, difference in y: -1 -3=-4. So squared differences: 9 and 16. Sum is 25. Square root is 5. Correct. So distance 5. Sum of radii √10 +1≈4.16. So 5 > 4.16. So separate circles. Therefore, 4 common tangent lines. So answer D.But let me check again. Let me compute √10 exactly. √10 is approximately 3.16227766. So 3.16227766 +1≈4.16227766. 5 is indeed greater. Therefore, circles are separate. So 4 common tangents.But wait, in some sources, they say that when circles are separate, there are 4 common tangents: two direct and two transverse. However, if the radii are such that one circle is much larger than the other, maybe the number reduces? But in standard Euclidean geometry, two circles not overlapping and not one inside the other have four common tangents regardless of their radii. So as long as distance > sum of radii, 4. If distance = sum, 3. If sum > distance > difference, 2. If distance = difference, 1. If distance < difference, 0.Therefore, according to that, answer should be D:4 lines.But wait, the problem is from an exam, maybe the options are different? Let me check again.Wait, original problem:"The lines that are tangent to both circle C₁: x²+y²+2x−6y=0 and circle C₂: x²+y²−4x+2y+4=0 are (　　)A: 1 line B: 2 lines C: 3 lines D: 4 lines"Yes, options are 1,2,3,4. So as per the analysis, answer should be D.But wait, let me make sure there is no miscalculation. Let me check radii and centers again.C₁: x²+y²+2x−6y=0. Completing the square:x² + 2x + y² -6y = 0 (x +1)^2 -1 + (y -3)^2 -9 =0 (x +1)^2 + (y -3)^2 =10. So center (-1,3), radius √10. Correct.C₂: x²+y²−4x+2y+4=0 x² -4x + y² +2y = -4 (x -2)^2 -4 + (y +1)^2 -1 = -4 (x -2)^2 + (y +1)^2 =1. So center (2,-1), radius 1. Correct.Distance between centers: between (-1,3) and (2,-1).Difference in x: 2 - (-1)=3 Difference in y: -1 -3=-4 Distance squared: 3² + (-4)^2 =9+16=25 Distance=5. Correct.Sum of radii: √10 +1≈3.16+1≈4.16. Distance is 5>4.16. So separate circles, 4 common tangents. Therefore, answer is D. So I think that's the correct answer. The answer should be D.But let me check another way. Maybe using the formula for the number of common tangents. The formula depends on the positions:Number of common tangents = - 4 if distance > sum of radii,- 3 if distance = sum,- 2 if sum > distance > |r1 - r2|,- 1 if distance = |r1 - r2|,- 0 if distance < |r1 - r2|.Here, |r1 - r2| = |√10 -1| ≈3.16 -1≈2.16. The distance between centers is 5, which is greater than sum of radii (~4.16), so 4 common tangents. Hence D.Therefore, I think the answer is D:4 lines.Final Answerboxed{D}</think>To determine the number of common tangent lines between the circles (C_1) and (C_2), we first convert their equations to standard form by completing the square.For circle (C_1: x^2 + y^2 + 2x - 6y = 0):- Group and complete the square for (x) and (y): [ (x^2 + 2x) + (y^2 - 6y) = 0 implies (x + 1)^2 - 1 + (y - 3)^2 - 9 = 0 implies (x + 1)^2 + (y - 3)^2 = 10 ]- Center: ((-1, 3))- Radius: (sqrt{10})For circle (C_2: x^2 + y^2 - 4x + 2y + 4 = 0):- Group and complete the square for (x) and (y): [ (x^2 - 4x) + (y^2 + 2y) = -4 implies (x - 2)^2 - 4 + (y + 1)^2 - 1 = -4 implies (x - 2)^2 + (y + 1)^2 = 1 ]- Center: ((2, -1))- Radius: (1)Next, calculate the distance between the centers:- Distance formula: [ sqrt{(2 - (-1))^2 + (-1 - 3)^2} = sqrt{(3)^2 + (-4)^2} = sqrt{9 + 16} = sqrt{25} = 5 ]Compare the distance with the sum and difference of the radii:- Sum of radii: (sqrt{10} + 1 approx 4.16)- Distance between centers: (5)Since the distance (5) is greater than the sum of the radii ((approx 4.16)), the circles are separate. Therefore, there are 4 common tangent lines.The answer is (boxed{D}).,<think>Okay, so I need to figure out how many common tangent lines there are between two circles, C₁ and C₂. The answer choices are 1, 2, 3, or 4 lines. Let me start by recalling what determines the number of common tangents between two circles. If two circles are separate, they have 4 common tangents. If they touch externally, they have 3. If they intersect at two points, they have 2. If one is entirely inside the other and they don't touch, they have 0. If they touch internally, they have 1. So first, I need to find the positions of these two circles relative to each other.First, let me write down the equations again:Circle C₁: x² + y² + 2x - 6y = 0 Circle C₂: x² + y² - 4x + 2y + 4 = 0I need to convert these equations into standard form by completing the squares.Starting with C₁:x² + 2x + y² - 6y = 0 Group the x terms and y terms:(x² + 2x) + (y² - 6y) = 0 Complete the square for x:Take coefficient of x, which is 2, divide by 2 to get 1, square it to get 1. Add and subtract 1.Similarly for y: coefficient of y is -6, divide by 2: -3, square it: 9. Add and subtract 9.So:(x² + 2x + 1 - 1) + (y² - 6y + 9 - 9) = 0 Which is:(x + 1)² - 1 + (y - 3)² - 9 = 0 Combine constants:(x + 1)² + (y - 3)² - 10 = 0 So:(x + 1)² + (y - 3)² = 10 Therefore, center of C₁ is (-1, 3) and radius is √10 ≈ 3.16.Now for C₂:x² - 4x + y² + 2y = -4 Again, group x and y terms:(x² - 4x) + (y² + 2y) = -4 Complete the squares:For x: coefficient of x is -4, divide by 2: -2, square it: 4. Add and subtract 4.For y: coefficient of y is 2, divide by 2: 1, square it: 1. Add and subtract 1.So:(x² - 4x + 4 - 4) + (y² + 2y + 1 - 1) = -4 Which becomes:(x - 2)² - 4 + (y + 1)² - 1 = -4 Combine constants:(x - 2)² + (y + 1)² - 5 = -4 So:(x - 2)² + (y + 1)² = 1 Therefore, center of C₂ is (2, -1) and radius is 1.Now, to find the positions of the circles, I need the distance between the centers and compare it to the sum and difference of the radii.First, calculate the distance between centers.Center of C₁: (-1, 3) Center of C₂: (2, -1)Use the distance formula:d = √[(2 - (-1))² + (-1 - 3)²] = √[(3)² + (-4)²] = √[9 + 16] = √25 = 5Sum of the radii: √10 + 1 ≈ 3.16 + 1 = 4.16 Difference of the radii: |√10 - 1| ≈ 3.16 - 1 = 2.16Now, compare the distance between centers (5) with the sum (≈4.16) and difference (≈2.16).Since 5 > 4.16, the circles are separate (i.e., they don't intersect and neither is inside the other). Therefore, according to my initial thought, there should be 4 common tangents. But wait, let me confirm again.Wait, the sum of the radii is approximately 4.16, and the distance between centers is 5, which is larger. So yes, the circles are externally separate, so 4 common tangents. Then the answer would be D: 4 lines.But wait, wait. Let me check again because sometimes when one circle is entirely inside another, but here the distance is 5, and the radii are √10 (~3.16) and 1. So if the distance between centers is 5, and each radius is 3.16 and 1. So the larger radius is about 3.16. So 3.16 + 1 = 4.16. The distance between centers is 5, which is more than 4.16, so indeed, the circles are separate. Therefore, they have four common tangents: two direct and two transverse.But let me check if the distance is greater than the sum of radii, so external tangents: 4. If the distance is equal to the sum, 3. If they intersect at two points, 2. If one is inside the other, 0. If they touch internally, 1. So since 5 > 4.16, so four tangents. So answer D: 4 lines.Wait, but the answer choices are A:1, B:2, C:3, D:4. So D is 4 lines. But the options given include 4 lines, which would be D. Hmm. But I need to make sure.Wait, but maybe I made a mistake in the calculation. Let me check again.Centers:C₁: (-1, 3)C₂: (2, -1)Difference in x: 2 - (-1) = 3Difference in y: -1 - 3 = -4Distance squared: 3² + (-4)² = 9 + 16 = 25. So distance is 5. Correct.Radius of C₁: √10 ≈ 3.16Radius of C₂: 1Sum of radii: √10 + 1 ≈ 4.16 < 5, so indeed, circles are separate, so four common tangents.Therefore, answer is D:4 lines.Wait, but let me check with another approach. If two circles are separate, there are 4 common tangents. If they are touching externally, 3. If intersecting, 2. If touching internally, 1. If one inside the other, 0. So in this case, since the distance is more than the sum, 4. So D.But let me think again. Maybe there is a mistake in the calculation. Wait, perhaps in the radius?Wait, C₁: x² + y² + 2x -6y =0. Converting to standard:(x +1)^2 + (y -3)^2 = 1 + 9 = 10, yes, radius sqrt(10). Correct.C₂: x² + y² -4x +2y +4 =0. Converting:(x -2)^2 + (y +1)^2 = 4 +1 -4 =1. Wait, 4 comes from (-4x completed as (x-2)^2 = x² -4x +4, so subtract 4; for y: 2y becomes (y+1)^2 -1, so subtract 1. So total equation:(x -2)^2 + (y +1)^2 -4 -1 +4 =0 → (x -2)^2 + (y +1)^2 -1 =0 → (x -2)^2 + (y +1)^2 =1. Correct. So radius 1.Distance between centers is 5, sum of radii is sqrt(10) +1 ≈4.16, which is less than 5, so separate. So four common tangents. So the answer is D. So why is the problem giving options up to D:4 lines. So yeah, D is the correct answer.But let me check once again. Maybe I misapplied the common tangent numbers. Wait, if two circles are separate, they have four common tangents: two direct (external) tangents and two transverse (internal) tangents. Wait, but when they are separate, the internal tangents cross between them, but if the circles are separate, the internal tangent lines would intersect the line connecting the centers between the circles. However, in this case, since the distance is greater than the sum of radii, the internal tangent lines do not exist. Wait, actually, no. Wait, internal and external tangents: when circles are separate, there are two external tangents (that don't cross between the circles) and two internal tangents (that cross between the circles). However, the internal tangents only exist when the circles are separate. If one circle is inside another, then there are no internal tangents. Wait, maybe I confused internal and external.Wait, let me check the definitions again.External tangent lines are the ones that do not cross the line segment joining the centers, while internal tangent lines cross the line segment. For two circles outside each other (distance > sum of radii), there are two external and two internal tangents, totaling four. If the circles are touching externally (distance = sum), then the two internal tangents become coincident, making three. If intersecting (distance between radii), two common tangents. If touching internally (distance = difference), one tangent. If one is inside the other, none.Hence, in this case, as the distance is greater than the sum, two external and two internal tangents, four in total. So the answer should be D:4 lines. Therefore, I think the correct answer is D. So the answer is D.Final Answerboxed{D}</think>To determine the number of common tangent lines between the circles ( C_1 ) and ( C_2 ), we first convert their equations into standard form by completing the square.For circle ( C_1: x^2 + y^2 + 2x - 6y = 0 ):- Group and complete the square for ( x ) and ( y ): [ (x + 1)^2 - 1 + (y - 3)^2 - 9 = 0 implies (x + 1)^2 + (y - 3)^2 = 10 ]- Center: ( (-1, 3) )- Radius: ( sqrt{10} approx 3.16 )For circle ( C_2: x^2 + y^2 - 4x + 2y + 4 = 0 ):- Group and complete the square for ( x ) and ( y ): [ (x - 2)^2 - 4 + (y + 1)^2 - 1 = -4 implies (x - 2)^2 + (y + 1)^2 = 1 ]- Center: ( (2, -1) )- Radius: ( 1 )Next, calculate the distance between the centers:- Distance formula: [ sqrt{(2 - (-1))^2 + (-1 - 3)^2} = sqrt{(3)^2 + (-4)^2} = sqrt{9 + 16} = sqrt{25} = 5 ]Compare the distance with the sum and difference of the radii:- Sum of radii: ( sqrt{10} + 1 approx 4.16 )- Difference of radii: ( |sqrt{10} - 1| approx 2.16 )Since the distance between the centers (5) is greater than the sum of the radii (4.16), the circles are separate. Therefore, there are 4 common tangent lines.[boxed{D}]]

🔑:To solve, we convert the equation of circle C_1: x^2+y^2+2x-6y=0 into its standard form, obtaining (x+1)^2+(y-3)^2=10. Therefore, the center of C_1 is (-1, 3), and its radius is r_1= sqrt{10}. Next, we convert the equation of circle C_2: x^2+y^2-4x+2y+4=0 into its standard form, obtaining (x-2)^2+(y+1)^2=1. Therefore, the center of C_2 is (2, -1), and its radius is r_2=1. The distance between the centers of the two circles is d= sqrt{(2+1)^2+(-1-3)^2}=5. Since r_1+r_2= sqrt{10}+1, we have d > r_1+r_2. Therefore, circles C_1 and C_2 are separate from each other. Thus, there are 4 common tangents to the two circles. Hence, the correct choice is: boxed{text{D}}. By converting the equations of the two circles into standard form, finding the centers and radii, and comparing the distance between the centers to the sum of the radii, we determine that the circles are separate, which means there are 4 common tangents. This problem examines the positional relationship between two circles, where the condition for two circles to be separate is that the distance between their centers is greater than the sum of their radii.

🧠:[<think>Okay, so I have this problem here with four functions, and I need to figure out which ones have the smallest positive period of π. The options are combinations of ① to ④, and the answer choices are A to D. Let me take it step by step.First, let me recall what the period of a function means. The period of a function is the smallest positive number T for which f(x + T) = f(x) for all x in the domain of f. So, I need to find the periods of each of these four functions and see which ones have π as their fundamental period.Let me list out the functions again:① y = sin |x|② y = |sin x|③ y = sin(2x + 2π/3)④ y = tan(2x + 2π/3)Starting with function ①: y = sin |x|Hmm, sin|x|. The absolute value affects the argument of the sine function. Normally, sin x has a period of 2π. But with |x|, the graph for x ≥ 0 is sin x, and for x < 0, it's sin(-x) = -sin x. Wait, but with absolute value, is it sin|x| or |sin x|? Here, it's sin|x|, so the argument is |x|. Let me think.For x positive, it's sin x, which is periodic with period 2π. For x negative, since |x| becomes positive, sin|x| would be sin(-x) when x is negative. Wait, no. |x| is always non-negative, so sin|x| for x negative is sin(-x) which is -sin x. Wait, no, sin|x| when x is negative is sin(|x|) = sin(-x) because |x| = -x when x is negative. Wait, actually, no. Wait, |x| is just the magnitude, so sin|x| when x is negative is sin(-x) but since -x is positive, it's sin(-x) = -sin x. So, for x positive, sin|x| = sin x, and for x negative, sin|x| = sin(-x) = -sin x.So, the function sin|x| is actually equal to sin x for x ≥ 0 and -sin x for x < 0. Wait, but that would make the function an odd function? Wait, no. Because if you replace x with -x, sin| -x | = sin|x|. So the function is actually even. Because sin|x| is the same as sin|x| whether x is positive or negative. Wait, no: Let me test x and -x.Take x = a (positive). Then sin|a| = sin a.Take x = -a. Then sin| -a | = sin a. So, sin|x| is actually sin|x| which is equal to sin x for x ≥ 0 and sin(-x) = sin x for x < 0. Wait, that's not right. Wait, sin|x| is just sin of the absolute value of x. So, for any x, whether positive or negative, sin|x| = sin(|x|). Since |x| is non-negative, this is equivalent to sin y where y = |x|. So, the graph would be the right half of the sine wave (for x ≥ 0) and the mirror image of the right half across the y-axis (for x < 0). So, instead of the usual sine wave that oscillates symmetrically around the origin, this would be a wave that is symmetric around the y-axis. So, for x ≥ 0, it's sin x, which has a period of 2π. For x < 0, it's the mirror image. So, the entire function sin|x| would have a period of 2π, right? Because after 2π, the right side repeats, and the left side is just the mirror image. Wait, but is there a smaller period?Wait, suppose we consider T as a period. Then sin|x + T| = sin|x| for all x. Let's test T = π. If T = π, then sin|x + π| should equal sin|x| for all x. Let's take x = 0: sin|π| = sin π = 0, and sin|0| = 0. So that works. But x = π/2: sin|π/2 + π| = sin|3π/2| = sin(3π/2) = -1, whereas sin|π/2| = 1. So, -1 ≠ 1, so T = π is not a period. Hmm, so that's a problem. Then maybe T = 2π is the period.Wait, but let me check x = π. sin|π + T| should equal sin|π|. Let’s take T = 2π. Then sin|π + 2π| = sin|3π| = sin 3π = 0, and sin|π| = 0. For x = π/2: sin|π/2 + 2π| = sin|5π/2| = sin(5π/2) = 1, and sin|π/2| = 1. For x negative, say x = -π/2: sin|-π/2 + 2π| = sin|3π/2| = sin(3π/2) = -1, but sin|-π/2| = sin(π/2) = 1. Wait, -1 ≠ 1. Wait, that can't be. Wait, no. Wait, when x is negative, say x = -a, a > 0. Then sin|x + T| = sin| -a + T |. If T = 2π, then sin| -a + 2π |. Since -a + 2π is positive (if a < 2π). So sin| -a + 2π | = sin(2π - a) = -sin a. But sin|x| at x = -a is sin| -a | = sin a. So, sin(2π - a) = -sin a. But that is not equal to sin a unless sin a = 0. So that's a problem. Therefore, even T = 2π is not a period?Wait, this is confusing. Maybe the period of sin|x| is not 2π? Let's think again.Alternatively, maybe the function sin|x| is not periodic at all? Wait, that seems contradictory because for x positive, it's sin x, which is periodic. But for x negative, it's also sin x, but mirrored. Wait, actually, if we consider x ≥ 0, sin|x| = sin x, which has period 2π. But if we consider the entire domain, because the function is even (sin|x| = sin| -x |), so it's symmetric about the y-axis. So, in order for the function to repeat, the entire graph from x = 0 to x = 2π must repeat. However, because the left side is just a mirror image of the right side, if we shift by 2π, the right side shifts to the right by 2π, but the left side is not shifting in the same way. Wait, maybe it's not periodic?Wait, hold on, this is getting complicated. Let me check specific points. Take x = 0. sin|0| = 0. If we shift by T, sin|0 + T| = sin T. For periodicity, we need sin T = 0, so T must be a multiple of π. Let's try T = π. Then sin|x + π| should equal sin|x| for all x.Take x = π/2: sin|π/2 + π| = sin|3π/2| = sin(3π/2) = -1. But sin|π/2| = 1. So -1 ≠ 1. So T = π is not a period.Take T = 2π: sin|x + 2π|. For any x, |x + 2π| = x + 2π if x + 2π ≥ 0, which it is for all x ≥ -2π. But since we are considering all real x, for x < -2π, |x + 2π| = -(x + 2π). But sin|x + 2π| = sin(x + 2π) for x ≥ -2π, which is sin x. For x < -2π, sin|x + 2π| = sin(-x - 2π) = sin(-x) = -sin x. But sin|x| for x < -2π is sin(-x) = -sin x. Wait, but sin|x + 2π| for x < -2π is sin(-x - 2π) = sin(-x - 2π) = sin(-x)cos(2π) - cos(-x)sin(2π) = -sin x. So sin|x + 2π| = -sin x for x < -2π. But sin|x| for x < -2π is sin(-x) = -sin x. Therefore, sin|x + 2π| = sin|x| for x < -2π?Wait, for x < -2π, let's take x = -3π. Then sin|x + 2π| = sin| -3π + 2π | = sin| -π | = sin π = 0. sin|x| = sin| -3π | = sin 3π = 0. So that works. Wait, but if x = -5π/2. Then sin|x + 2π| = sin| -5π/2 + 2π | = sin| -π/2 | = sin(π/2) = 1. sin|x| = sin| -5π/2 | = sin(5π/2) = 1. So that works. Wait, maybe T = 2π is a period?Wait, but earlier when I took x = -π/2 and T = 2π, sin|x + T| = sin| -π/2 + 2π | = sin| 3π/2 | = sin(3π/2) = -1, but sin|x| at x = -π/2 is sin| -π/2 | = sin(π/2) = 1. So here, sin|x + T| = -1, which is not equal to sin|x| = 1. Therefore, T = 2π is not a period? Wait, that contradicts the previous examples.Wait, but x = -π/2 is in the interval x ≥ -2π, right? Because x = -π/2 is greater than -2π. So, for x ≥ -2π, |x + 2π| = x + 2π. Therefore, sin|x + 2π| = sin(x + 2π) = sin x. But sin|x| for x ≥ 0 is sin x, and for x < 0, sin|x| = sin(-x) = -sin x. Wait, no. Wait, for x between -2π and 0, x is negative, so |x| = -x. So sin|x| = sin(-x) = -sin x. However, sin|x + 2π| for x between -2π and 0 is sin(x + 2π). Since x + 2π is between 0 and 2π, so sin(x + 2π) = sin x. But sin|x| = -sin x. So unless sin x = -sin x, which is only when sin x = 0, these two are not equal. So, for x in (-2π, 0), sin|x + 2π| = sin x + 2π = sin x, but sin|x| = sin(-x) = -sin x. Therefore, sin x ≠ -sin x unless sin x = 0. Therefore, T = 2π is not a period for sin|x|. Therefore, the function sin|x| is not periodic? That seems strange.But when we plot sin|x|, for x ≥ 0, it's the standard sine wave, and for x < 0, it's the mirror image, which is also a sine wave but flipped. So from x = -∞ to ∞, it's like two sine waves glued at x = 0, one on the right going up and down, and one on the left mirroring it. But in that case, there's no periodicity because the left side is a mirror image, not a translation. So maybe sin|x| is not periodic? Therefore, it doesn't have a period? But the question is about functions with the smallest positive period of π. If the function isn't periodic, then it doesn't have a period, so it can't be one of the answers. But the options include ① in some answers (A and B). So maybe my reasoning is wrong.Wait, maybe I need to consider that for x ≥ 0, sin|x| = sin x, which is periodic with period 2π. But for the entire function, maybe the period is still 2π because the mirror image part would repeat every 2π as well? Wait, but when you shift by 2π, the right side shifts to the right, but the left side shifts to the right as well, so the left side would now be from x = -2π to 0, but originally, the left side was a mirror image of the right side. So shifting right by 2π would make the left side (from x = -2π to 0) be a copy of sin x from x = -2π to 0, which is sin x shifted left by 2π, so sin(x + 2π) = sin x. Wait, but sin|x + 2π| = sin(x + 2π) for x + 2π ≥ 0, which is x ≥ -2π, and sin|x + 2π| = sin(-x - 2π) for x + 2π < 0. So, for x ≥ -2π, sin|x + 2π| = sin(x + 2π) = sin x. But sin|x| for x ≥ 0 is sin x, and for x between -2π and 0, sin|x| = sin(-x) = -sin x. Therefore, sin|x + 2π| = sin x for x between -2π and 0, but sin|x| = -sin x. So unless sin x = -sin x, which is only when sin x = 0, this doesn't hold. Therefore, sin|x + 2π| ≠ sin|x| for all x. Therefore, sin|x| is not periodic. That seems to be the case.But then the answer options include ① in some choices. So maybe I made a mistake here.Wait, let me check another source or think differently. Maybe the period is π?Wait, if I take T = π. Then sin|x + π| versus sin|x|. Let's take x = 0: sin|π| = sin π = 0. sin|0| = 0. That's okay. x = π/2: sin|π/2 + π| = sin(3π/2) = -1. sin|π/2| = 1. Not equal. So T = π is not a period. x = π: sin|2π| = 0, sin|π| = 0. Okay. x = -π/2: sin| -π/2 + π | = sin(π/2) = 1. sin| -π/2 | = sin(π/2) = 1. Okay here. x = -π: sin| -π + π | = sin 0 = 0, sin| -π | = sin π = 0. Okay. x = -3π/2: sin| -3π/2 + π | = sin| -π/2 | = sin(π/2) = 1. sin| -3π/2 | = sin(3π/2) = -1. So here, sin|x + π| = 1 ≠ sin|x| = -1. Therefore, T = π is not a period.Alternatively, maybe the period is something else, but given that shifting by T makes some points equal and others not, it seems that sin|x| is not periodic. Therefore, it doesn't have a period. So, among the four functions, ① does not have period π. So the answer can't include ①, so options C and D are left. Wait, but original options are A: all four, B: ①②③, C: ②③④, D: ②③.But if ① isn't periodic, then the answer can't be A or B. So then it's between C and D. Let me check the other functions.Function ②: y = |sin x|The absolute value of sin x. The period of sin x is 2π, but taking absolute value reflects the negative parts to positive, so the period halves. So |sin x| has period π. Let me verify.For |sin(x + π)| = |sin x cos π + cos x sin π| = | -sin x | = |sin x|. So yes, T = π is a period. And since the original period was 2π, and we're reflecting, there's no smaller period than π. So ② has period π. So ② is good.Function ③: y = sin(2x + 2π/3)The general sine function y = sin(Bx + C) has period 2π / |B|. Here, B = 2, so period is 2π / 2 = π. Therefore, period is π. So ③ has period π. So ③ is good.Function ④: y = tan(2x + 2π/3)The tangent function has period π / |B| for y = tan(Bx + C). Here, B = 2, so period is π / 2. Wait, π / 2? Wait, but let me confirm. The standard period of tan x is π. So for tan(Bx + C), the period is π / |B|. So here, B = 2, so period is π / 2. So the period of ④ is π / 2, which is smaller than π. But the question is asking for functions with the smallest positive period of π. So if the period is π / 2, then π is not the smallest period. Therefore, ④ does not have period π.Wait, but hold on. Let me double-check. The period of tan(2x + 2π/3) is π / 2. Because tan repeats every π, so if the argument is 2x + 2π/3, then the function will repeat when the argument increases by π. So 2x + 2π/3 + π = 2(x + π/2) + 2π/3. Therefore, replacing x with x + π/2 gives the same value. So the period is π/2. Therefore, the period is π/2, so the smallest period is π/2, not π. So ④ does not have period π. Therefore, the functions with period π are ② and ③. Therefore, the answer is D: ②③.Wait, but the options are:A: ①②③④B: ①②③C: ②③④D: ②③So based on my reasoning:①: Not periodic (or if periodic, period is undefined or larger than π), so doesn't have period π.②: Period π.③: Period π.④: Period π/2.Therefore, only ② and ③ have period π. So the correct answer is D.But wait, let me check again for function ①. Maybe I was wrong earlier. Let's take another approach.Function ①: y = sin|x|. Let's check its graph.For x ≥ 0, it's sin x. For x < 0, it's sin(-x) = -sin x. Wait, no. Wait, sin|x| when x < 0 is sin(-x) = sin(-x) = -sin x. Wait, but sin|x| is sin of |x|. So for x < 0, |x| = -x, so sin|x| = sin(-x) = -sin x. Wait, so the graph for x < 0 is -sin x. Wait, but that would make the function odd. However, sin|x| is actually an even function because sin| -x | = sin|x|. But according to this, for x positive, sin|x| = sin x, and for x negative, sin|x| = sin(-x) = -sin x. Wait, that would make the function even?Wait, no. Let's plug in x and -x. For x positive, f(-x) = sin|-x| = sin x. For x positive, f(x) = sin x. So f(-x) = sin x = f(x). So it's even. Wait, but sin(-x) = -sin x. Wait, here sin|x| is sin of |x|, which is always non-negative. So for x positive, sin|x| = sin x. For x negative, sin|x| = sin(-x) = sin x. Wait, because |x| is just the magnitude. So sin|x| = sin x for x positive and sin x for x negative. Wait, that can't be. Wait, no. If x is negative, |x| = -x (since x is negative), so sin|x| = sin(-x) = -sin x. Wait, no. Wait, sin(-x) = -sin x, but |x| is positive. So sin|x| for x negative is sin(|x|) = sin(-x) = -sin x. Wait, but sin(-x) is -sin x, so sin|x| = -sin x when x is negative? But that would make sin|x| an odd function. But sin|x| is even, because |x| is even. Wait, let me verify with x = a and x = -a.Take x = a: sin|a| = sin a.Take x = -a: sin|-a| = sin a. Therefore, sin|x| is an even function. Wait, so sin|x| for x negative is sin(|x|) = sin(a) where a = |x|. So regardless of x being positive or negative, sin|x| = sin(|x|). Therefore, for x negative, sin|x| = sin(-x) only if you consider x negative. Wait, this is confusing. Let's take a concrete example.Let x = -π/2. Then sin|x| = sin(π/2) = 1. Similarly, x = π/2: sin|x| = sin(π/2) = 1. So for both x = π/2 and x = -π/2, sin|x| = 1. Therefore, the function is even. Therefore, the graph is symmetric about the y-axis.So for x ≥ 0, it's the standard sine wave starting at 0, going up to 1 at π/2, back to 0 at π, down to -1 at 3π/2, back to 0 at 2π, etc. For x < 0, it mirrors the x ≥ 0 part. So the graph for x < 0 is the reflection of x > 0 across the y-axis. Therefore, from x = -2π to 0, it's a mirror image of x = 0 to 2π.So, to check periodicity, we need to see if shifting by some T will map the graph onto itself. For example, if we take T = 2π, shifting the entire graph to the right by 2π. But since the graph is mirrored on the left side, shifting right by 2π would not align the left side. However, if we shift by π, let's see:Take the graph from x = 0 to π: it's sin x from 0 to π. The graph from x = π to 2π is sin x from π to 2π. For the negative side, x = -π to 0 is the mirror image of x = 0 to π. So if we shift by π, shifting right by π, the part from x = 0 to π would move to x = π to 2π, which is sin x from π to 2π, which is the negative part. However, the original graph at x = π to 2π is sin x from π to 2π, which is negative, but shifted right by π, x = π to 2π would come from x = 0 to π, which is positive. So they don't match. Hence, shifting by π doesn't work.Alternatively, shifting by 2π. The right side shifts right by 2π, but the left side shifts right into the region from x = -2π to -0, which was originally a mirror image of x = 0 to 2π. But after shifting, the part from x = 0 to 2π becomes x = 2π to 4π, and the part from x = -2π to 0 shifts to x = 0 to 2π. However, the original graph from x = 0 to 2π is sin x, and the shifted left part (original -2π to 0) is the mirror image, which when shifted becomes 0 to 2π but as a mirror image, i.e., sin x for x from 0 to 2π but mirrored. Wait, this is getting too complicated. Maybe it's better to accept that sin|x| is not periodic because the left side is a mirror image and not a translation. Therefore, it doesn't repeat periodically. Therefore, ① is not periodic. Hence, the answer is D: ②③.But wait, let me check online or think again. Maybe I can recall that the function sin|x| is not periodic. For example, references indicate that sin|x| is not a periodic function because the left side doesn't repeat the right side's behavior through translation. So yes, ① is not periodic. So functions ② and ③ have period π, ④ has period π/2, ① is not periodic. Therefore, the answer is D: ②③.Final Answerboxed{D}</think>To determine which functions have the smallest positive period of ( pi ), we analyze each function:1. Function ( y = sin |x| ): - The function ( sin |x| ) is even and consists of the right half of the sine wave for ( x geq 0 ) and its mirror image for ( x < 0 ). - Testing periodicity shows that shifting by ( pi ) or ( 2pi ) does not satisfy the periodicity condition for all ( x ). Therefore, ( sin |x| ) is not periodic.2. Function ( y = |sin x| ): - The absolute value of ( sin x ) reflects the negative parts of the sine wave, halving its period. - The period of ( |sin x| ) is ( pi ).3. Function ( y = sin(2x + frac{2pi}{3}) ): - The general period of ( sin(Bx + C) ) is ( frac{2pi}{B} ). Here, ( B = 2 ), so the period is ( pi ).4. Function ( y = tan(2x + frac{2pi}{3}) ): - The general period of ( tan(Bx + C) ) is ( frac{pi}{B} ). Here, ( B = 2 ), so the period is ( frac{pi}{2} ).Thus, the functions with the smallest positive period of ( pi ) are ( ② ) and ( ③ ).[boxed{D}],<think>Okay, so I need to figure out which of these four functions have the smallest positive period of π. The options are given as A, B, C, D, each listing different combinations of the functions ① to ④. Let me start by recalling what the period of a function means. The period is the smallest positive number T for which f(x + T) = f(x) for all x. So I need to determine the period of each function and see which ones have π as their period.First, let's list out the functions again to be clear:① y = sin|x|② y = |sin x|③ y = sin(2x + 2π/3)④ y = tan(2x + 2π/3)I need to check each of these functions one by one.Starting with ①: y = sin|x|Hmm, sin|x|. Let me think. The absolute value inside the sine function might affect its periodicity. Normally, sin(x) has a period of 2π. But when we take sin|x|, the function is no longer the same as sin x for negative x. For x ≥ 0, sin|x| = sin x, which is periodic with period 2π. But for x < 0, sin|x| = sin(-x) = -sin x. Wait, but if we take absolute value inside, so sin|x| would actually be sin|x|, which for x negative is sin(-x) = -sin x. Wait, but sin|x| is equal to sin(|x|), so for x positive, it's sin x, and for x negative, it's sin(-x) = -sin x. Wait, but that's the same as sin|x|, right? Because |x| is always non-negative. So sin|x| is equivalent to sin(|x|). So for x positive, it's sin x, and for x negative, it's sin(-x) = -sin x. Wait, no, sin|x| is not -sin x for x negative. Because |x| is positive, so sin|x| is equal to sin(-x) when x is negative, which is -sin x? Wait, hold on. Wait, sin(-x) = -sin x, but sin|x| when x is negative is sin(-x) = -sin x. Wait, but the function y = sin|x| would be: for x ≥ 0, sin x; for x < 0, sin(-x) = -sin x. So this is actually an odd function, because sin|x| is equal to sin x for x positive and -sin x for x negative. Wait, but that's the same as the sign function times sin|x|. Wait, actually, no. Let me graph it mentally. For x positive, it's the regular sine wave. For x negative, it's sin(-x) which is -sin x, so actually, it's a mirror image of the sine function across the y-axis for the negative side. Wait, that might create a different periodicity. Let me check. Let's test if sin|x| is periodic with period 2π. Let's take x = 0. Then sin|0| = 0. If we take x = π, sin|π| = sin π = 0. Then x = π + T, but I need to check if sin|x + T| = sin|x| for all x. But because of the absolute value, adding T to x might not preserve the periodicity. Let's suppose T is the period. Let's test T = π. Let's see: for x positive, sin|x + π| = sin(x + π) = -sin x. But sin|x| = sin x. So that's not equal. So period cannot be π. What about T = 2π? For x positive, sin(x + 2π) = sin x, which works. For x negative, say x = -a where a > 0, sin|x + 2π| = sin(-x + 2π) if x + 2π is positive. Wait, if x is negative, say x = -b, b > 0, then x + 2π = 2π - b. So sin|x + 2π| = sin(2π - b) = -sin b. But sin|x| for x = -b is sin b. So sin|x + 2π| = -sin b ≠ sin|x|. Therefore, T = 2π is not a period. Wait, so this is a problem. So maybe sin|x| is not periodic at all? Wait, that can't be. Let me think again. Wait, when x is negative, sin|x| = sin(-x) = -sin x. So for example, if we take x = -π/2, sin|x| = sin(π/2) = 1. Then x + 2π = -π/2 + 2π = 3π/2, sin|x + 2π| = sin(3π/2) = -1. But sin|x| at x = -π/2 is 1, but at x + 2π, it's -1. So that's not equal. Therefore, the function is not periodic? Wait, but how can that be? Maybe I'm misunderstanding. Let me check.Wait, actually, the function sin|x| is not periodic. Because for the positive side, it behaves like sin x, which is periodic with 2π, but for the negative side, it's a reflection. However, since the negative side is a reflection of the positive side, but when you shift the function, the negative side doesn't align with the positive side's periodicity. So perhaps sin|x| is not a periodic function. Wait, but when I look at x from 0 to infinity, it's sin x, and from negative infinity to 0, it's sin(-x) = -sin x. So the graph would be: for positive x, the usual sine wave, and for negative x, it's the sine wave inverted. So this creates a function that is symmetric about the y-axis, but with the left side inverted. But periodicity requires that shifting the function by T results in the same graph. However, because the left and right sides are different, shifting by T would not reproduce the same graph. Therefore, perhaps sin|x| is not periodic. Therefore, it does not have a period. But the problem states "the smallest positive period". Hmm. Wait, maybe I need to confirm this.Alternatively, maybe I made a mistake. Let me check some references. Wait, since I can't actually look things up, I need to think through. Let's consider specific values. Let's take x = π/2. sin|π/2| = sin(π/2) = 1. Now, if we add a period T, then sin|π/2 + T| should equal 1. Similarly, for x = -π/2, sin| -π/2 | = sin(π/2) = 1. Then adding T to x = -π/2 gives -π/2 + T. The absolute value is | -π/2 + T |. Depending on T, this could be different. Suppose T = π. Then x + T = -π/2 + π = π/2. So sin| -π/2 + π | = sin(π/2) = 1. So that works. But what about x = π. sin|π| = sin π = 0. Then x + T = π + π = 2π. sin|2π| = sin(2π) = 0. So that works. What about x = π/4. sin|π/4| = sin(π/4) = √2/2. Then x + T = π/4 + π = 5π/4. sin|5π/4| = sin(5π/4) = -√2/2. But sin|π/4| is √2/2, and sin|5π/4| is -√2/2. These are not equal. Therefore, T = π does not work. So perhaps T is not π.Wait, but maybe another T? Let me test T = 2π. Take x = π/4. Then x + T = π/4 + 2π = 9π/4. sin|9π/4| = sin(9π/4) = sin(π/4) = √2/2. Which matches sin|π/4|. For x = -π/2, x + T = -π/2 + 2π = 3π/2. sin|3π/2| = sin(3π/2) = -1. But sin|x| at x = -π/2 is sin(π/2) = 1. So sin|3π/2| = -1 ≠ 1. So T = 2π also doesn't work. Hmm. Therefore, maybe the function sin|x| is not periodic. If that's the case, then it doesn't have a period. But the question is about the functions with the smallest positive period of π. If the function isn't periodic, then it doesn't have a period, so it can't have a period of π. Therefore, ① does not have period π.But wait, maybe I'm missing something. Let me think again. The function sin|x| is even, right? Because sin|−x| = sin|x|. So it's an even function. So if it's periodic, the period would have to satisfy T such that sin|x + T| = sin|x| for all x. But because of the absolute value, the periodicity would have to work for both positive and negative x. Let's suppose T is a period. Then for x > 0, sin(x + T) = sin x. So T must be 2πn, where n is integer. Similarly, for x < -T, sin|x + T| = sin(-x - T) = sin(-x - T) = sin(-(x + T)) = -sin(x + T). But since the function is even, sin|x + T| should equal sin|x|. For x negative, say x = -a, a > 0. Then sin|x + T| = sin| -a + T |. If T is 2πn, then sin| -a + 2πn |. Depending on the value of a and n, this may or may not equal sin| -a | = sin a. For example, take n = 1, T = 2π. Then sin| -a + 2π | = sin(2π - a) = -sin a. But sin| -a | = sin a. So -sin a ≠ sin a unless sin a = 0. Therefore, unless a is multiple of π, this doesn't hold. Therefore, T = 2π is not a period. Therefore, the function sin|x| is not periodic. Therefore, ① does not have a period of π. So ① is out.Moving on to ②: y = |sin x|Okay, the absolute value of sin x. The regular sin x has period 2π, but when we take the absolute value, the negative parts become positive. So the graph is the same as sin x for the positive parts, and flipped for the negative parts. So the period becomes π instead of 2π. Let's verify that. The standard period of |sin x| is π. Let's check if |sin(x + π)| = |sin x|. Let's compute |sin(x + π)| = | - sin x | = |sin x|. Yes, that's correct. Therefore, the period is π. So ② has a period of π.Next, ③: y = sin(2x + 2π/3)For a sine function of the form sin(Bx + C), the period is 2π / |B|. Here, B is 2. Therefore, the period is 2π / 2 = π. So the period is π. So ③ has a period of π.④: y = tan(2x + 2π/3)Tangent functions have a period of π / |B| for functions of the form tan(Bx + C). Here, B is 2, so the period is π / 2. Wait, but π / 2 is smaller than π. So the period of ④ is π / 2. Therefore, the period is π / 2, which is smaller than π. But the question is asking for functions with the smallest positive period of π. Therefore, ④ has a smaller period than π, so it doesn't qualify. So ④ is out.So summarizing:①: Not periodic (or no period), so doesn't have period π.②: Period π.③: Period π.④: Period π/2.Therefore, the functions with the smallest positive period of π are ② and ③, which is option D.Wait, but let me double-check ① again. The original question is in Chinese, translated to English. The function is y = sin|x|. Maybe there's a different interpretation? For example, sin(|x|) versus sin|x|. But in Chinese, maybe the notation is different? Wait, no. In LaTeX, it's written as y = sin |x|, which is sin(|x|). So as we discussed earlier. So if it's not periodic, then it doesn't have a period. So ① is not considered.But let me check once more. Suppose I consider x >= 0, sin|x| is sin x, which is periodic with period 2π. For x < 0, sin|x| is sin(-x) = -sin x. So if we shift x by 2π, for x positive, sin(x + 2π) = sin x, but for x negative, sin|x + 2π| = sin(-x + 2π) = sin(2π - x) = -sin x. But sin|x| for x negative is -sin x, so sin|x + 2π| = -sin x = sin|x|. Wait, so perhaps when x is negative, sin|x + 2π| = sin(2π - x) = sin(2π - x) = -sin x. But sin|x| when x is negative is sin(-x) = -sin x. So sin|x + 2π| = -sin x = sin|x|. So actually, for x negative, shifting by 2π gives the same value. For x positive, shifting by 2π gives the same value. Therefore, T = 2π is a period. Wait, but earlier when I tested x = -π/2, T = 2π would give x + T = 3π/2. Then sin|3π/2| = sin(3π/2) = -1. But sin|x| at x = -π/2 is sin(π/2) = 1. Wait, that contradicts. Wait, if x = -π/2, then sin|x| = sin(π/2) = 1. Then x + 2π = -π/2 + 2π = 3π/2, sin|x + 2π| = sin(3π/2) = -1. So 1 vs -1. Not equal. Therefore, T = 2π is not a period. Therefore, this suggests that the function is not periodic. Therefore, my initial conclusion was correct. Therefore, ① does not have period 2π or π. So it's not periodic, so it's out.Therefore, the answer should be D: ②③. So among the options given, D is the correct one.Final Answerboxed{D}</think>To determine which functions have the smallest positive period of ( pi ), we analyze each function:1. ( y = sin |x| ): - This function is not periodic. For ( x geq 0 ), it behaves like ( sin x ), and for ( x < 0 ), it behaves like ( -sin x ). Testing periodicity shows that it does not satisfy ( sin |x + T| = sin |x| ) for any ( T ), hence it is not periodic.2. ( y = |sin x| ): - The absolute value of ( sin x ) changes the period from ( 2pi ) to ( pi ). This is because ( |sin(x + pi)| = |-sin x| = |sin x| ), confirming the period is ( pi ).3. ( y = sin(2x + frac{2pi}{3}) ): - The general period of ( sin(Bx + C) ) is ( frac{2pi}{B} ). Here, ( B = 2 ), so the period is ( frac{2pi}{2} = pi ).4. ( y = tan(2x + frac{2pi}{3}) ): - The general period of ( tan(Bx + C) ) is ( frac{pi}{B} ). Here, ( B = 2 ), so the period is ( frac{pi}{2} ), which is smaller than ( pi ).Thus, the functions with the smallest positive period of ( pi ) are ② and ③. The correct answer is (boxed{D}).]

🔑:AnalysisThis question examines the smallest positive period of trigonometric functions, which can be solved by evaluating each function individually.SolutionFor the function (y=sin |x|), it is not a periodic function, thus it does not meet the condition.Let (y=f(x)=|sin x|), then (f(x+π)=|sin (x+π)|=|-sin x|=|sin x|=f(x)), (therefore) the function (y=|sin x|) has the smallest positive period of (π), which meets the condition; Furthermore, the function (y=sin (2x+ dfrac{2π}{3}) ) has the smallest positive period of (T= dfrac{2π}{2}=π ), which meets the condition; The function (y=tan (2x+ dfrac{2π}{3}) ) has the smallest positive period of (T= dfrac{π}{2} ), which does not meet the condition.In summary, among the four functions, the ones with the smallest positive period of (π) are (②③). Therefore, the correct choice is boxed{text{D}}.

🧠:[<think>Okay, so I need to find the sum of the leading digits of the 2nd, 3rd, 4th, 5th, and 6th roots of a number N. N is a 313-digit number where every digit is 7. The answer choices are given, so I need to figure out each f(r) for r from 2 to 6 and then add them up.First, let me understand what the leading digit of the rth root means. If I take the rth root of N, it's going to be some number, and the leading digit is the first digit of that number. For example, if the 2nd root (square root) of N is 777...777 (but obviously smaller), the first digit would be 7 or something else. But since N is a very large number with all 7s, I need a way to approximate its roots.Since N is a 313-digit number, all 7s, it's equal to 7 times 10^0 + 7 times 10^1 + ... + 7 times 10^312. But that's a bit complicated. Maybe I can approximate N as 7 times 10^312, but that's just the first digit. Wait, no. Actually, a number with 313 digits all 7s is 777...777 (313 times). So it's actually 7*(10^313 - 1)/9. Because 777...777 (n digits) is equal to 7*(10^n - 1)/9. Let me check that: For example, 777 = 7*100 + 7*10 + 7*1 = 7*(100 + 10 + 1) = 7*(111) = 7*(10^3 - 1)/9. Yes, since 111 = (10^3 -1)/9. So generalizing, a number with n digits all 7 is 7*(10^n -1)/9. So N = 7*(10^313 -1)/9.But when taking roots, maybe it's easier to approximate N as roughly 7*10^312, since (10^313 -1)/9 is approximately 10^313/9, and multiplying by 7 gives 7*10^313/9 ≈ 7.777...*10^312. So N is approximately 7.777...*10^312. So log10(N) is approximately log10(7.777...*10^312) = log10(7.777...) + 312. Let me compute log10(7.777...). Since 7.777... is equal to 7 + 7/9 = 70/9 ≈ 7.777777...So log10(70/9) = log10(70) - log10(9) ≈ 1.8451 - 0.9542 = 0.8909. Therefore, log10(N) ≈ 0.8909 + 312 = 312.8909.Therefore, the rth root of N is N^(1/r) = 10^(log10(N)/r) = 10^(312.8909/r). Let me compute 312.8909 divided by r for each r from 2 to 6. Then, the leading digit can be found by looking at the fractional part of that quotient, which when exponentiated with base 10 gives a number between 1 and 10, and the integer part of that exponentiated value is the leading digit.Wait, maybe I need to recall that if log10(N) = k + f, where k is integer and 0 ≤ f < 1, then N = 10^{k + f} = 10^k * 10^f, so the leading digit is floor(10^f). So similarly, when we take log10(N)/r = m + g, where m is integer and 0 ≤ g < 1, then N^(1/r) = 10^{m + g} = 10^m * 10^g, so leading digit is floor(10^g). So yes, the fractional part of log10(N)/r determines the leading digit.Therefore, for each r, I need to compute (312.8909)/r, separate the integer part and the fractional part, then compute 10^{fractional part}, and take the integer part of that to get the leading digit.So let's compute log10(N)/r for each r from 2 to 6.First, log10(N) ≈ 312.8909.For r=2: 312.8909 / 2 = 156.44545So the fractional part is 0.44545. Then 10^{0.44545} ≈ 10^{0.445} ≈ ?We know that 10^{0.4771} ≈ 3, since log10(3) ≈ 0.4771. Similarly, 10^{0.447} is closer to sqrt(10)/2 or something? Let me recall that 10^{0.3} ≈ 2, 10^{0.3010} ≈ 2, 10^{0.447} = ?Alternatively, maybe use natural logs. Let me compute ln(10^{0.44545}) = 0.44545 * ln(10) ≈ 0.44545 * 2.302585 ≈ 1.025. Then e^{1.025} ≈ e^1 * e^{0.025} ≈ 2.718 * 1.0253 ≈ 2.718 * 1.0253 ≈ 2.718 + 2.718*0.0253 ≈ 2.718 + 0.0688 ≈ 2.7868. So 10^{0.44545} ≈ 2.7868. Therefore, the leading digit is 2. So f(2)=2.Wait, but let me confirm with better approximation. Alternatively, maybe use known logarithm values.We know that log10(2) ≈ 0.3010log10(3) ≈ 0.4771log10(2.5) = log10(5/2) = log10(5) - log10(2) ≈ 0.69897 - 0.3010 ≈ 0.39797log10(2.7) = ?Let me compute 10^{0.44545}.Alternatively, maybe use linear approximation between 10^{0.4} and 10^{0.5}.10^{0.4} ≈ 2.5118864315095810^{0.45} = sqrt(10^{0.4} * 10^{0.5}) ≈ sqrt(2.511886 * 3.162277) ≈ sqrt(7.943) ≈ 2.818But 0.44545 is 0.45 - 0.00455, so maybe 10^{0.44545} ≈ 10^{0.45} / 10^{0.00455} ≈ 2.818 / (1 + 0.00455*ln(10)) ≈ 2.818 / (1 + 0.0105) ≈ 2.818 / 1.0105 ≈ 2.789. So approximately 2.789. So leading digit is 2. Therefore, f(2)=2. Hmm.Wait, but maybe my approximation is off. Let me check with more accurate method.Alternatively, since 10^{0.445} = e^{0.445*ln10} ≈ e^{0.445*2.302585} ≈ e^{1.02465} ≈ 2.786. So yes, about 2.786, leading digit 2.Wait, but in the answer choices, the possible sum is up to 29, so maybe the leading digits are higher. Wait, perhaps my approximation is wrong. Let's check more carefully.Wait, log10(N) is approximately 312.8909. Let's confirm that first. Since N is a 313-digit number, all 7s, so log10(N) is between log10(7*10^312) and log10(8*10^312). log10(7*10^312) = log10(7) + 312 ≈ 0.8451 + 312 = 312.8451. log10(8*10^312) = log10(8) + 312 ≈ 0.9031 + 312 = 312.9031. Since N is 777...777, which is 7*(10^313 -1)/9. So log10(N) = log10(7) + log10(10^313 -1) - log10(9). But 10^313 -1 ≈ 10^313, so log10(N) ≈ log10(7) + 313 - log10(9). log10(7) ≈ 0.8451, log10(9) ≈ 0.9542. So log10(N) ≈ 0.8451 + 313 - 0.9542 ≈ 313 - 0.1091 ≈ 312.8909. So that's correct.Therefore, log10(N) ≈ 312.8909.So for each root r, we compute log10(N)/r, which is approximately 312.8909/r. Let's compute that.Starting with r=2: 312.8909 / 2 = 156.44545. So this is 156 + 0.44545. Then 10^{0.44545} ≈ as above, ~2.786, leading digit 2. So f(2)=2.r=3: 312.8909 /3 ≈ 104.2969666. So the fractional part is 0.2969666. Then 10^{0.29697} ≈ ?We know that log10(2) ≈ 0.3010, which is very close to 0.29697. So 10^{0.29697} ≈ 10^{0.3010 - 0.00403} ≈ 2 / 10^{0.00403} ≈ 2 / (1 + 0.00403*ln10) ≈ 2 / (1 + 0.00928) ≈ 2 / 1.00928 ≈ 1.9816. Therefore, approximately 1.9816, which is just under 2. So the leading digit is 1? Wait, but 1.98 is still leading digit 1? Wait, no. Wait, 1.98 is between 1 and 10, but the leading digit is the first digit, which is 1 here. Wait, but 1.98 is almost 2, but since it's less than 2, the leading digit is 1. But is that correct?Wait, hold on. The leading digit is the first digit. So if N^(1/3) is approximately 1.98... * 10^104, then the number is 1.98... followed by 104 zeros, so the leading digit is 1. But 1.98 is close to 2, but still less than 2. Wait, but is this accurate?Alternatively, maybe my approximation is off. Let me check 10^{0.29697} more accurately.Let me compute 0.29697 * ln10 ≈ 0.29697 * 2.302585 ≈ 0.6835. Then e^{0.6835} ≈ e^{0.6835} ≈ 1 + 0.6835 + (0.6835)^2/2 + (0.6835)^3/6 + ... Hmm, maybe tedious. Alternatively, use known that e^{0.6835} ≈ e^{0.6931 - 0.0096} ≈ e^{0.6931} * e^{-0.0096} ≈ 2 * (1 - 0.0096) ≈ 2 * 0.9904 ≈ 1.9808. So yes, e^{0.6835} ≈ 1.9808. Therefore, 10^{0.29697} ≈ 1.9808, so approximately 1.9808, which is 1.98... So leading digit is 1. Therefore, f(3)=1.Wait, but 1.98 is very close to 2, but since it's less than 2, the leading digit is 1. Hmm, but maybe my approximation is wrong?Alternatively, maybe use logarithms. If 10^{0.29697} = x, then log10(x) = 0.29697. So x = 10^{0.29697}. Let's use logarithm tables or remember that 10^{0.3010} = 2. So 0.29697 is 0.3010 - 0.00403. So 10^{0.29697} = 10^{0.3010 - 0.00403} = 2 / 10^{0.00403}. 10^{0.00403} ≈ 1 + 0.00403 * ln10 ≈ 1 + 0.00403 * 2.302585 ≈ 1 + 0.00928 ≈ 1.00928. Therefore, 2 / 1.00928 ≈ 1.9816. So x ≈ 1.9816, which is 1.98, leading digit 1. Therefore, f(3)=1. Hmm.Wait, but maybe the actual value is slightly over 1.98, which is still leading digit 1. So, okay. Let's tentatively say f(3)=1.Wait, but is there a mistake here? Because N is a 313-digit number, so when taking the cube root, we're getting a number with about 313/3 ≈ 104.3 digits, which would be a 105-digit number? Wait, no. Wait, the number of digits of N^(1/r) is floor(log10(N)/r) +1. Since log10(N^(1/r)) = log10(N)/r ≈ 312.8909/r. For r=3, that's 104.2969..., so the number is 10^{104.2969} = 10^{0.2969} * 10^{104}, which is a 105-digit number (since 10^{104} is 1 followed by 104 zeros, so multiplying by 10^{0.2969} which is ~1.98 gives a 105-digit number). But the leading digit is 1.98, so first digit is 1. Therefore, f(3)=1.But I need to make sure. Let me check if my approximation of log10(N) is correct. Since N is (10^313 -1)/9 *7, log10(N) = log10(7) + log10(10^313 -1) - log10(9). But since 10^313 -1 is very close to 10^313, so log10(10^313 -1) ≈ 313 - log10(e)/10^313 ≈ 313, negligible. Therefore, log10(N) ≈ log10(7) + 313 - log10(9) ≈ 0.8451 + 313 - 0.9542 ≈ 312.8909. That's correct.Therefore, the computation for r=3 is correct. So f(3)=1.Moving on to r=4: log10(N)/4 ≈ 312.8909 /4 ≈ 78.222725. So fractional part is 0.222725. Then 10^{0.222725} ≈ ?We know that log10(1.666) ≈ 0.2218, so 10^{0.2218} ≈ 1.666. Similarly, 10^{0.2227} is slightly higher. Let me compute 10^{0.2227}.Again, using natural logs: 0.2227 * ln10 ≈ 0.2227 * 2.302585 ≈ 0.5126. Then e^{0.5126} ≈ 1.669. Alternatively, use the fact that 10^{0.2227} = e^{0.2227*ln10} ≈ e^{0.5126} ≈ 1.669. So approximately 1.669. Therefore, leading digit is 1. So f(4)=1.Wait, but 1.669 is leading digit 1. Hmm. Wait, 1.669 is 1.6... so first digit 1. Yes.Alternatively, 10^{0.2227} ≈ 1.669. So leading digit 1.Therefore, f(4)=1.r=5: log10(N)/5 ≈ 312.8909 /5 ≈ 62.57818. Fractional part 0.57818. Then 10^{0.57818} ≈ ?What's 10^{0.57818}? Let's recall that 10^{0.57} is approx 3.715, since log10(3.715) ≈ 0.57. Wait, 10^{0.57} = ?Alternatively, let's compute 0.57818*ln10 ≈ 0.57818*2.302585 ≈ 1.331. Then e^{1.331} ≈ 3.78. Because e^1 = 2.718, e^1.1 ≈ 3.004, e^1.2≈3.320, e^1.3≈3.669, e^1.331≈3.78. So 10^{0.57818} ≈ 3.78. Therefore, leading digit is 3. Therefore, f(5)=3.Wait, but maybe check more accurately. log10(3.7) ≈ 0.5682, log10(3.8)≈0.5798. So 0.57818 is between log10(3.7) and log10(3.8). Let's compute log10(3.78) ≈ ?3.78: log10(3.78) = log10(3) + log10(1.26) ≈ 0.4771 + 0.1004 ≈ 0.5775. So log10(3.78) ≈ 0.5775. But our exponent is 0.57818, which is slightly higher. So 10^{0.57818} ≈ 3.78 * 10^{0.00068} ≈ 3.78 * 1.0016 ≈ 3.786. So approximately 3.786, leading digit 3. Therefore, f(5)=3.Okay, moving on to r=6: log10(N)/6 ≈ 312.8909 /6 ≈ 52.1484833. So fractional part is 0.1484833. Then 10^{0.14848} ≈ ?We know that log10(1.4) ≈ 0.1461, log10(1.41) ≈ 0.1491. So 0.14848 is between log10(1.4) and log10(1.41). Let's interpolate. Let me compute 10^{0.14848}.Let me use natural logs again: 0.14848*ln10 ≈ 0.14848*2.302585 ≈ 0.3419. Then e^{0.3419} ≈ 1.407. Therefore, 10^{0.14848} ≈ 1.407. Therefore, leading digit is 1. Therefore, f(6)=1.Alternatively, using log10(1.407) ≈ 0.1481, so 0.14848 is slightly higher. So 10^{0.14848} ≈ 1.407 + (0.14848 - 0.1481)*(1.41 -1.407)/ (0.1491 -0.1481) ≈ 1.407 + (0.00038)*(0.003)/0.001 ≈ 1.407 + 0.00114 ≈ 1.40814. So approximately 1.408, leading digit 1. So f(6)=1.So adding them up: f(2) + f(3) + f(4) + f(5) + f(6) = 2 + 1 + 1 + 3 +1 = 8. But the answer choices are: (A)8, (B)9, etc. So the answer would be 8? But let's check again because maybe my approximations were off.Wait, but let's verify each step again, because 8 is one of the options, but I have to be careful here.First, for r=2: log10(N)/2=156.44545. So fractional part is 0.44545. 10^{0.44545}≈2.786. Leading digit 2. Correct.r=3: fractional part 0.29697. 10^{0.29697}≈1.9816. Leading digit 1. Correct.r=4: fractional part 0.2227. 10^{0.2227}≈1.669. Leading digit 1. Correct.r=5: fractional part 0.57818. 10^{0.57818}≈3.78. Leading digit 3. Correct.r=6: fractional part 0.14848. 10^{0.14848}≈1.407. Leading digit 1. Correct.So sum is 2+1+1+3+1=8. So answer is A)8. But wait, the options have 8 as (A). But I need to check if my initial approximation of log10(N) as 312.8909 is correct. Let me verify that.N is a 313-digit number with all 7s. So N = 7*(10^{313}-1)/9. Let's compute log10(N):log10(N) = log10(7) + log10(10^{313} -1) - log10(9). Since 10^{313} -1 ≈ 10^{313}, so log10(10^{313} -1) ≈ 313. Therefore, log10(N) ≈ log10(7) + 313 - log10(9). Let me compute that exactly:log10(7) ≈ 0.84509804log10(9) ≈ 0.954242509Thus, log10(N) ≈ 0.84509804 + 313 - 0.954242509 ≈ 313 - 0.109144469 ≈ 312.8908555. So my initial approximation was correct, log10(N)≈312.8908555. So accurate.Therefore, when dividing by r=2,3,4,5,6, the fractional parts are correctly calculated. Therefore, the leading digits as 2,1,1,3,1. Sum is 8. So answer is A)8.But wait, the options have (A)8, (B)9, etc. So 8 is an option. But I need to check if there's any error in the calculation.Wait, let's check r=5 again. log10(N)/5 ≈62.57818. So fractional part 0.57818. 10^{0.57818}≈3.78. So leading digit 3. Is this correct?Alternatively, maybe use more accurate calculation for 10^{0.57818}. Let me compute 0.57818*ln10 ≈0.57818*2.302585≈1.331. e^{1.331} ≈3.78. Yes. Alternatively, use Taylor series.But maybe it's better to check if log10(3.78)=log10(3)+log10(1.26)=0.4771+0.1004≈0.5775. So log10(3.78)=0.5775. So to get log10(x)=0.57818, x=3.78*10^{0.00068}≈3.78*1.0016≈3.786. So x≈3.786. Leading digit is 3. Correct.Similarly, for r=2, 0.44545, 10^{0.44545}≈2.786. So leading digit 2.Wait, but let me check if there's a different approach. Maybe using the fact that N ≈7.777...*10^{312}, so log10(N)≈log10(7.777...)+312.But log10(7.777...) is log10(70/9)=log10(70)-log10(9)=1.8451 -0.9542=0.8909, as before. So log10(N)=0.8909 +312=312.8909. So same as previous.Alternatively, perhaps we can model N as 7.777...*10^{312}, so N ≈ (7.777...)*10^{312}. Therefore, when taking the rth root, N^{1/r} ≈ (7.777...)^{1/r}*10^{312/r}. The leading digit would be the first digit of (7.777...)^{1/r}*10^{312/r}.But 10^{312/r} is 10^{integer + fractional} = 10^{integer}*10^{fractional}, so the leading digit is determined by (7.777...)^{1/r}*10^{fractional part}. Wait, but how is this different?Wait, maybe this approach is similar. Let me check for r=2. Then N^(1/2) ≈ sqrt(7.777...)*10^{156}. sqrt(7.777...) is approx 2.786, as before. So leading digit 2.Similarly, for r=3: cube root of 7.777... is approx 1.98, leading digit 1.Wait, so actually, (7.777...)^{1/r} is the key factor here. For each r, compute (70/9)^{1/r} and the leading digit is determined by that number times 10^{k}, where k is such that when multiplied by 10^{312/r}, the decimal is adjusted. Wait, maybe this is the same as before.Alternatively, perhaps we can compute log10(N) = log10(70/9) + 312 = (log10(70) - log10(9)) + 312 ≈ (1.8451 -0.9542) + 312 ≈0.8909 +312≈312.8909. So same as before. So regardless of the approach, the calculation remains the same.Therefore, the leading digits are 2,1,1,3,1. Sum is 8. Therefore, answer is A)8.But wait, let me check if someone else has done this problem before. Wait, this is a problem similar to past AMC/AIME problems. The key idea is using logarithms to find leading digits. So the steps seem correct.Alternatively, maybe the question is tricky in some way. For example, maybe the approximation N ≈7.777...*10^312 is not precise enough? But since N is 313 digits, all 7s, the difference between N and 7.777*10^312 is 7.777*10^312 - N = 7.777*10^312 - 7*(10^313 -1)/9. Let me compute that:7*(10^313 -1)/9 = (7*10^313 -7)/9 = 7*10^313/9 - 7/9 ≈7.777...*10^312 -0.777...Therefore, N is exactly 7.777...*10^312 -0.777..., but when taking log10(N), the difference is negligible because 0.777... is much smaller than 7.777...*10^312. So log10(N) ≈ log10(7.777...*10^312) = log10(7.777...) +312≈0.8909 +312≈312.8909. So that's correct.Therefore, the leading digits computed are accurate. So sum is 8.But the answer choices include 29, which is much higher. Wait, but maybe I made a mistake in taking the fractional part. Wait, when I divide log10(N) by r, which is 312.8909/r, and separate into integer and fractional parts. For example, with r=2: 312.8909/2=156.44545, so fractional part is 0.44545, which is correct.Similarly for r=3: 312.8909/3≈104.2969666, fractional part 0.2969666. Correct.r=4: 78.222725, fractional 0.222725. Correct.r=5:62.57818, fractional 0.57818. Correct.r=6:52.1484833, fractional 0.1484833. Correct.Therefore, fractional parts correctly calculated. Then, converting each fractional part to 10^{fraction}, leading digit as floor(10^{fraction}).But 10^{0.44545}=2.786, floor is 2.10^{0.29697}=1.98, floor is 1.10^{0.2227}=1.669, floor is1.10^{0.57818}=3.78, floor is3.10^{0.14848}=1.407, floor is1.Therefore, sum 2+1+1+3+1=8.Therefore, answer is A)8. So I think that's correct. Maybe the answer is 8, option A.But let me check with an exact example. Suppose we take N as 7777 (4-digit number), then compute roots and leading digits.For example, N=7777. Then log10(N)=log10(7777)≈3.8903. Then sqrt(N)=88.19..., leading digit 8.Wait, but according to our method, log10(N)=3.8903. Divide by 2:1.94515. Fractional part 0.94515. 10^{0.94515}≈8.81, leading digit 8. Which matches the actual square root of 7777≈88.19, leading digit 8. So the method works here.Similarly, cube root of 7777≈19.76, leading digit 1. Let's check with log10(N)/3≈3.8903/3≈1.29677. Fractional part 0.29677. 10^{0.29677}≈1.98, leading digit 1. Which matches 19.76≈19.76, leading digit1.Similarly, 4th root of 7777≈9.34, leading digit9. Wait, log10(N)/4≈0.972575. So fractional part is0.972575. 10^{0.972575}≈9.38, leading digit9. Which matches.But according to our method for N=7777, f(4)=9. But according to my previous problem for r=4, we had fractional part 0.2227 and leading digit1. Wait, but in this small example, the fractional part was high, leading to high leading digit. So why in the original problem, for r=4, we get a low leading digit?Ah, because in the original problem, log10(N) is 312.8909, which when divided by 4 gives 78.2227. The fractional part is0.2227, which gives 10^{0.2227}≈1.669, leading digit1. But in the smaller N=7777, the fractional part after dividing log10(N) by4 was0.9725, leading to high leading digit. So the difference is because in the original problem, the log10(N)/r is a number with a large integer part and small fractional part, leading to 10^{small fractional} between1 and10. In the smaller N, the log10(N)/r has a fractional part close to1, leading to higher leading digit.Therefore, the method is correct. So I think in the original problem, the sum is8. Therefore, answer isA)8. But wait, the answer choices include (C)11, (D)22, (E)29. So why is 8 an option? Maybe the question is not from an official source, but the answer is indeed8. But let me check once more.Wait, let me test with r=1. log10(N)/1=312.8909. Fractional part0.8909. 10^{0.8909}=7.777..., which matches the leading digit7 of N. So the method works here.Therefore, the method is validated. So in r=2 to6, the leading digits are2,1,1,3,1. Sum8. So answer isA)8.But maybe the problem is different. Wait, let me check the original problem again. "Let N be the positive integer 7777…777, a 313-digit number where each digit is a 7. Let f(r) be the leading digit of the rth root of N. What is f(2) + f(3) + f(4) + f(5)+ f(6)?"Yes, that's the problem. So my calculation seems correct. Therefore, the answer should be8, optionA. But the system might expect a different answer. Wait, let me check if someone else has done this problem.Alternatively, perhaps the error is in the assumption that N is approximately7.777*10^312. But N is exactly (10^313 -1)/9*7. So log10(N)=log10(7)+(313)-log10(9) - log10(1 -10^{-313}). Since10^{-313} is negligible, so log10(N)=log10(7) - log10(9) +313. So log10(N)=log10(7/9)+313≈log10(0.777...)+313≈-0.1091+313=312.8909. So same as before.Alternatively, maybe taking into account the exact value of log10(N). Let me compute log10(7/9)=log10(7)-log10(9)=0.84509804 -0.954242509≈-0.109144469. Therefore, log10(N)=313 -0.109144469≈312.8908555.Therefore, when computing log10(N)/r, for r=2,3,4,5,6, the exact fractional parts:For r=2:312.8908555 /2=156.44542775. So fractional part is0.44542775.10^{0.44542775}=?As before, approximately2.786. Leading digit2.For r=3:312.8908555 /3=104.2969518. Fractional part0.2969518.10^{0.2969518}=~1.9816. Leading digit1.r=4:312.8908555 /4=78.222713875. Fractional0.222713875.10^{0.222713875}=1.669. Leading digit1.r=5:312.8908555 /5=62.5781711. Fractional0.5781711.10^{0.5781711}=3.78. Leading digit3.r=6:312.8908555 /6=52.148475916. Fractional0.148475916.10^{0.148475916}=1.407. Leading digit1.So no change from before. Still sum8.Therefore, the answer is8, optionA.Final Answerboxed{A}</think>Given ( N ), a 313-digit number where each digit is 7, we need to find the sum of the leading digits of the 2nd, 3rd, 4th, 5th, and 6th roots of ( N ).First, approximate ( N ) as ( 7.777 times 10^{312} ). The logarithm base 10 of ( N ) is:[log_{10}(N) approx log_{10}(7.777) + 312 approx 0.8909 + 312 = 312.8909]For each root ( r ), we compute ( log_{10}(N)/r ) and use the fractional part to determine the leading digit.1. For ( r = 2 ): [ frac{312.8909}{2} = 156.44545 quad text{(fractional part is 0.44545)} ] [ 10^{0.44545} approx 2.786 quad text{(leading digit is 2)} ] ( f(2) = 2 )2. For ( r = 3 ): [ frac{312.8909}{3} approx 104.29697 quad text{(fractional part is 0.29697)} ] [ 10^{0.29697} approx 1.9816 quad text{(leading digit is 1)} ] ( f(3) = 1 )3. For ( r = 4 ): [ frac{312.8909}{4} approx 78.2227 quad text{(fractional part is 0.2227)} ] [ 10^{0.2227} approx 1.669 quad text{(leading digit is 1)} ] ( f(4) = 1 )4. For ( r = 5 ): [ frac{312.8909}{5} approx 62.57818 quad text{(fractional part is 0.57818)} ] [ 10^{0.57818} approx 3.78 quad text{(leading digit is 3)} ] ( f(5) = 3 )5. For ( r = 6 ): [ frac{312.8909}{6} approx 52.14848 quad text{(fractional part is 0.14848)} ] [ 10^{0.14848} approx 1.407 quad text{(leading digit is 1)} ] ( f(6) = 1 )Summing these leading digits:[f(2) + f(3) + f(4) + f(5) + f(6) = 2 + 1 + 1 + 3 + 1 = 8]Thus, the answer is (boxed{A}).,<think>Okay, so I need to figure out this problem where N is a 313-digit number, all of which are 7s. Then, for each root from the 2nd to the 6th, I have to find the leading digit of that root and add them up. The answer choices are given, so that might help if I get stuck. Let me start by understanding the problem step by step.First, let me confirm what N is. It's a number with 313 digits, all 7s. So N is 777...777 with 313 sevens. That's a massive number! Now, for each r in {2,3,4,5,6}, we need to compute the rth root of N and find the leading digit of that root. Then sum those leading digits. The key here is probably logarithms, since leading digits relate to the mantissa of the logarithm. Remember that if a number x has leading digit d, then d × 10^k ≤ x < (d+1) × 10^k for some integer k. Taking the logarithm (base 10) of x, we get log10(x) = k + log10(d) + something, where 0 ≤ something < log10((d+1)/d). So the leading digit d can be found by looking at 10^{fractional part of log10(x)} and seeing which interval it falls into between d and d+1.So, for each root r, the rth root of N is N^(1/r). Let's denote x = N^(1/r). Then log10(x) = (1/r) * log10(N). If I can compute log10(N), then multiply by 1/r, and then look at the fractional part to find the leading digit.Therefore, the plan is:1. Compute log10(N)2. For each r in {2,3,4,5,6}, compute (1/r) * log10(N)3. Take the fractional part of each result and compute 10^{fractional part} to get the leading digit.But how do I compute log10(N)? N is a 313-digit number where each digit is 7. So N is between 7×10^312 and 8×10^312. Therefore, log10(N) is between log10(7×10^312) and log10(8×10^312). Calculating log10(7×10^312) = log10(7) + 312 ≈ 0.84509804 + 312 = 312.84509804Similarly, log10(8×10^312) = log10(8) + 312 ≈ 0.903089987 + 312 = 312.903089987But since all digits are 7, maybe we can get a better approximation. For example, N is 7.777...777 × 10^312. But how do we express that exactly?Wait, N is 7 repeated 313 times. So it's 7*(10^0 + 10^1 + 10^2 + ... + 10^312). That sum is a geometric series: (10^313 - 1)/9. Therefore, N = 7*(10^313 - 1)/9. Hmm, that's an exact expression. But how does that help me compute log10(N)?Alternatively, maybe approximate N as 7*(10^313)/9, since subtracting 1 might be negligible for such a large exponent. So N ≈ (7/9)*10^313. Then log10(N) ≈ log10(7/9) + 313 ≈ log10(0.777...) + 313. log10(7/9) is log10(7) - log10(9) ≈ 0.8451 - 0.9542 ≈ -0.1091. Therefore, log10(N) ≈ -0.1091 + 313 = 312.8909. So approximately 312.8909.But is this a good enough approximation? Since we're dealing with leading digits, small errors in log10(N) can affect the fractional part, which in turn affects the leading digit. So perhaps we need a more accurate estimation.Alternatively, let's note that N is a number with 313 digits all 7. So the first digit is 7, followed by 312 7s. So N is 7*10^312 + 7*10^311 + ... + 7*10^0. Which is 7*(10^313 - 1)/9, as I mentioned before.So exact value is N = (7/9)*(10^313 - 1). Then log10(N) = log10(7/9) + log10(10^313 - 1). But log10(10^313 - 1) is approximately 313 - log10(1 - 10^{-313}) ≈ 313 + (10^{-313}/ln(10)), which is negligible. So essentially, log10(N) ≈ log10(7/9) + 313 ≈ (log10(7) - log10(9)) + 313 ≈ (0.84509804 - 0.9542425094) + 313 ≈ (-0.1091444694) + 313 ≈ 312.8908555306. So log10(N) ≈ 312.8908555306.Therefore, log10(N) is approximately 312.8908555306. Let's store this as a precise value. Let me note that log10(7) ≈ 0.84509804, log10(9) ≈ 0.9542425094, so log10(7/9) ≈ -0.1091444694, yes.Therefore, for each r in {2,3,4,5,6}, compute (1/r) * log10(N) ≈ (312.8908555306)/r.Then, the integer part of this is going to be the exponent, and the fractional part is what determines the leading digit. Since when you take 10^{log10(x)} = x, so if log10(x) = k + f, where k is integer and 0 ≤ f < 1, then x = 10^{k + f} = 10^k * 10^f. So the leading digit is the integer part of 10^f.Therefore, for each r, we compute (312.8908555306)/r, subtract the integer part to get the fractional part, compute 10^{fractional part}, and take the integer part of that as the leading digit.So let's compute this step by step for each r.Starting with r=2:log10(N^(1/2)) = (1/2)*log10(N) ≈ 312.8908555306 / 2 ≈ 156.4454277653So the integer part is 156, fractional part is 0.4454277653Compute 10^{0.4454277653}10^0.4454277653 ≈ ?We can use logarithm properties or remember that 10^{0.447} ≈ 2.8, since log10(2.8) ≈ 0.447.Wait, 10^{0.4454} ≈ ?Let me compute 0.4454277653.We can note that 10^{0.4454} = e^{ln(10)*0.4454} ≈ e^{2.302585093*0.4454} ≈ e^{1.024} ≈ 2.784.Alternatively, since we know that log10(2.78) ≈ 0.444, log10(2.79) ≈ 0.4456. So 10^{0.4454} is approximately 2.79.Wait, let me check:log10(2.7) ≈ 0.4314log10(2.8) ≈ 0.4472So 0.4454 is between 2.7 and 2.8, closer to 2.8. Let me do a linear approximation.The difference between 0.4472 and 0.4314 is 0.0158 for a difference of 0.1 in the number (2.8 - 2.7 = 0.1). So 0.4454 is 0.4454 - 0.4314 = 0.014 above 2.7's log. So 0.014 / 0.0158 ≈ 0.886 of the interval. Therefore, 2.7 + 0.886*0.1 ≈ 2.7 + 0.0886 ≈ 2.7886. So approximately 2.7886. So leading digit is 2, but wait, 2.7886's integer part is 2. Wait, but 2.7886 is approximately 2.79, so leading digit is 2. Therefore, f(2) = 2.Wait, but is it possible that 10^{0.4454} is 2.78, which starts with 2, so yes, leading digit is 2. So f(2)=2.Wait, but wait, actually, the leading digit is the first digit of the number. So if x is N^(1/2), then x is approximately 2.78... × 10^156, so leading digit is 2. So f(2)=2. Hmm.But let me check with more precise calculation.Alternatively, since 10^{0.4454277653}:Let me compute ln(10^{0.4454277653}) = 0.4454277653 * ln(10) ≈ 0.4454277653 * 2.302585093 ≈ 1.025.Then e^{1.025} ≈ e^1 * e^0.025 ≈ 2.718 * 1.0253 ≈ 2.718 * 1.0253 ≈ 2.718 + 2.718*0.0253 ≈ 2.718 + 0.0687 ≈ 2.7867. So approx 2.787. So leading digit is 2. Therefore, f(2)=2.Okay, moving on to r=3.log10(N^(1/3)) = (1/3)*log10(N) ≈ 312.8908555306 / 3 ≈ 104.2969518435Integer part 104, fractional part 0.2969518435Compute 10^{0.2969518435}Again, 0.29695 is approx log10(1.975) because log10(2) = 0.3010, so 0.29695 is just slightly less than 0.3010. So 10^{0.29695} ≈ 1.975?Alternatively, compute 10^{0.29695}:Take natural logarithm again: ln(10^{0.29695}) = 0.29695 * ln(10) ≈ 0.29695 * 2.302585 ≈ 0.6835Then e^{0.6835} ≈ e^{0.6835} ≈ 1.981 (since e^0.7 ≈ 2.013, so subtract a bit). Let's compute:e^0.6835 ≈ e^{0.6 + 0.0835} = e^0.6 * e^0.0835 ≈ 1.8221 * 1.087 ≈ 1.8221 * 1.087 ≈ approx 1.980. So approximately 1.98. Therefore, leading digit is 1. Therefore, f(3)=1? Wait, but 1.98 is almost 2.0. So why isn't the leading digit 1? Wait, 1.98 is between 1 and 2, but since it's 1.98, the leading digit is 1.But wait, 1.98 is just 1.98..., so the first digit is 1. Hmm, but if it's 1.98, then the leading digit is 1. Wait, but if you have a number like 1.98 x 10^something, the leading digit is 1. Wait, but maybe my approximation is off.Wait, maybe 10^{0.29695} is actually 1.98, which would round to 2.0 when considering precision? Let me check with exact logarithm.log10(1.98) ≈ 0.2967. So 10^{0.29695} ≈ 1.98. Therefore, it's approximately 1.98, so leading digit is 1. So f(3)=1.Wait, but is that correct? If the number is 1.98... then the leading digit is 1. However, 1.98 is very close to 2.0, but since the number is just less than 2.0, the leading digit is 1. Wait, but perhaps due to the approximation, maybe the actual value is slightly higher?Wait, let's calculate 10^{0.2969518435} more precisely.We can use the Taylor series expansion around a known point. Let's take x = 0.2969518435. Let's note that 10^{0.2969518435} = e^{ln(10)*x} ≈ e^{2.302585093*x}. So x ≈ 0.2969518435Multiply 2.302585093 * 0.2969518435 ≈ 2.302585093 * 0.2969518435Let me compute:2.302585093 * 0.2969518435First, 2 * 0.29695 = 0.59390.302585093 * 0.2969518435 ≈ 0.302585 * 0.2969518 ≈ approx 0.09 (exact: 0.302585 * 0.2969518 ≈ 0.302585 * 0.3 ≈ 0.0907755, but since it's 0.29695, subtract 0.003048*0.302585 ≈ 0.000921, so ≈ 0.0907755 - 0.000921 ≈ 0.0898545)Therefore, total is ≈ 0.5939 + 0.0898545 ≈ 0.6837545Therefore, e^{0.6837545} ≈ e^{0.6837545}We know e^0.683 ≈ 1.980 (since e^0.683 ≈ 1.980). Let me check:Compute 0.6837545We can use e^0.6837545 = e^{0.6 + 0.0837545} = e^0.6 * e^{0.0837545}e^0.6 ≈ 1.822118800e^{0.0837545} ≈ 1 + 0.0837545 + (0.0837545)^2/2 + (0.0837545)^3/6Compute:First term: 1Second term: 0.0837545 ≈ 0.08375Third term: (0.08375)^2 / 2 ≈ 0.007015625 / 2 ≈ 0.0035078125Fourth term: (0.08375)^3 / 6 ≈ 0.000586914 / 6 ≈ 0.000097819Adding these: 1 + 0.08375 + 0.0035078125 + 0.000097819 ≈ 1.087355631Therefore, e^{0.0837545} ≈ 1.087355631Multiply by e^0.6 ≈ 1.8221188 * 1.0873556 ≈ Let's compute 1.8221188 * 1 = 1.8221188, 1.8221188 * 0.0873556 ≈ approx 1.8221188 * 0.08 = 0.1457695, 1.8221188 * 0.0073556 ≈ approx 0.01342. So total ≈ 0.1457695 + 0.01342 ≈ 0.1591895. Then total ≈ 1.8221188 + 0.1591895 ≈ 1.9813083. Therefore, e^{0.6837545} ≈ 1.9813. Therefore, 10^{0.2969518435} ≈ 1.9813. So approximately 1.9813. Therefore, the leading digit is 1. But wait, 1.9813 is very close to 2.0. But since it's still less than 2.0, the leading digit remains 1. However, this is extremely close. Is there a chance that my approximation of log10(N) might be slightly off, which would tip this over to 2.0?Wait, let's recall that log10(N) was approximated as 312.8908555306. But is that exact? N is (7/9)(10^313 - 1). So log10(N) = log10(7) - log10(9) + log10(10^313 - 1). Since 10^313 -1 is very close to 10^313, log10(10^313 -1 ) ≈ 313 - epsilon, where epsilon is tiny. So log10(N) ≈ log10(7) - log10(9) + 313 - epsilon. So log10(N) ≈ 313 + (log10(7) - log10(9)) - epsilon. Since log10(7) - log10(9) ≈ -0.1091444694, so log10(N) ≈ 313 - 0.1091444694 - epsilon ≈ 312.8908555306 - epsilon. Therefore, our original approximation is actually an overestimate, because we neglected the epsilon. So log10(N) is slightly less than 312.8908555306.Therefore, when we computed (1/3)*log10(N) ≈ 104.2969518435, but actually it's slightly less. So fractional part is slightly less than 0.2969518435. Let's say fractional part is 0.29695 instead of 0.2969518435. Then 10^{0.29695} ≈ 1.9813 as before. If the fractional part is slightly less, say 0.2969, then 10^{0.2969} is slightly less. Let me compute 10^{0.2969}:Similarly, log10(1.98) ≈ 0.2967. So 0.2969 is 0.0002 more. So 10^{0.2967} = 1.98, 10^{0.2969} = 1.98 * 10^{0.0002} ≈ 1.98 * (1 + 0.0002 * ln10) ≈ 1.98 * (1 + 0.0004605) ≈ 1.98 + 0.000911 ≈ 1.980911. So approximately 1.9809. So even if the fractional part is 0.2969, the value is still ~1.9809. Therefore, leading digit is 1. So even if we take into account the slight underestimation, the leading digit is still 1. Therefore, f(3)=1.Hmm, but the answer choices don't include 1 as part of the sum. Wait, the options are 8,9,11,22,29. So if we have f(2)=2, f(3)=1, then we need to check the rest. But maybe my initial calculations are wrong. Let's check r=3 again.Wait, maybe there's a mistake in the approximation. Let's try a different approach. Let's use more accurate logarithm values.We had log10(N) ≈ 312.8908555306. Let's use more decimal places. Let me get log10(7) and log10(9) to more decimal places.log10(7) ≈ 0.8450980400142568log10(9) ≈ 0.9542425094393249Therefore, log10(7/9) = log10(7) - log10(9) ≈ 0.8450980400142568 - 0.9542425094393249 ≈ -0.1091444694250681Therefore, log10(N) = log10(7/9) + log10(10^313 - 1) ≈ -0.1091444694250681 + (313 - log10(1 + 1/(10^313 - 1))) ≈ 312.8908555305749 (since log10(10^313 -1 ) ≈ 313 - ε, where ε is negligible). So log10(N) ≈ 312.8908555305749Therefore, when divided by 3:312.8908555305749 / 3 ≈ 104.29695184352496So fractional part is 0.29695184352496Compute 10^{0.29695184352496}Let me use the identity that 10^{x} = e^{x ln10}So x = 0.29695184352496ln(10) ≈ 2.302585093So x * ln10 ≈ 0.29695184352496 * 2.302585093 ≈ 0.683755Therefore, e^{0.683755} ≈ ?We can use the Taylor series again. Let's compute e^0.683755.Alternatively, use a calculator-like approach. Let's note that e^0.683755 is approximately equal to:We know that e^0.683755 is between e^0.68 and e^0.684.Compute e^0.68:e^0.6 = 1.822118800e^0.08 = 1.08328706767Therefore, e^0.68 = e^0.6 * e^0.08 ≈ 1.8221188 * 1.08328706767 ≈ 1.8221188 * 1.08 = 1.967, plus 1.8221188 * 0.003287 ≈ ~0.006, so total ≈ 1.973. So e^0.68 ≈ 1.973Similarly, e^0.684:Compute e^0.684. Let's do e^0.6 * e^0.084 ≈ 1.8221188 * e^0.084Compute e^0.084:Using Taylor series around 0.08:e^0.08 ≈ 1 + 0.08 + 0.08^2/2 + 0.08^3/6 ≈ 1 + 0.08 + 0.0032 + 0.0003413 ≈ 1.0835413Then e^0.084 ≈ e^0.08 * e^0.004 ≈ 1.0835413 * (1 + 0.004 + 0.004^2/2 + 0.004^3/6) ≈ 1.0835413 * (1.004 + 0.000008 + 0.000000042) ≈ 1.0835413 * 1.004008042 ≈ 1.0835413 + 1.0835413*0.004008 ≈ 1.0835413 + 0.004343 ≈ 1.0878843Therefore, e^0.684 ≈ 1.8221188 * 1.0878843 ≈ approx 1.8221188*1.08 = 1.967, plus 1.8221188*0.0078843 ≈ 0.01436, so total ≈ 1.967 + 0.01436 ≈ 1.98136Therefore, e^0.683755 is approximately halfway between 1.973 and 1.98136. Wait, but 0.683755 is 0.68 + 0.003755, so e^0.683755 = e^{0.68 + 0.003755} = e^0.68 * e^0.003755 ≈ 1.973 * (1 + 0.003755 + 0.000007) ≈ 1.973 * 1.003755 ≈ 1.973 + 1.973*0.003755 ≈ 1.973 + 0.00741 ≈ 1.9804.Therefore, e^{0.683755} ≈ 1.9804. Therefore, 10^{0.29695184352496} ≈ 1.9804, so the value is approximately 1.9804, which is 1.9804... So leading digit is 1. Therefore, f(3)=1.Wait, but 1.9804 is just under 2.0, so the leading digit is 1. So the third root of N is approximately 1.9804 × 10^104, so the leading digit is 1. Therefore, f(3)=1. Hmm. But let's check with another approach.Alternatively, if we use the exact log10(N) ≈ 312.8908555305749Then log10(N^(1/3)) = 312.8908555305749 / 3 ≈ 104.29695184352496The fractional part is 0.2969518435249610^{0.29695184352496} ≈ 1.9804 as above. So leading digit is 1. Therefore, f(3)=1.Wait, but maybe I need to check if the approximation of log10(N) is accurate enough. Let's see. Since N = (7/9)(10^313 - 1), then log10(N) = log10(7/9) + log10(10^313 - 1). But log10(10^313 - 1) = 313 + log10(1 - 10^{-313}). Then log10(1 - x) ≈ -x/ln(10) for small x. Therefore, log10(1 - 10^{-313}) ≈ -10^{-313}/ln(10). Therefore, log10(N) = log10(7/9) + 313 - 10^{-313}/ln(10). So the error in our approximation is approximately -10^{-313}/ln(10), which is negligible—practically zero. Therefore, our log10(N) is accurate to the decimal places we used. Therefore, 10^{0.29695184352496} ≈ 1.9804 is correct, leading digit 1. Therefore, f(3)=1.But wait, according to the answer choices, the possible sums are 8,9,11,22,29. If f(2)=2, f(3)=1, then even if the rest are higher digits, like 3,4,5, the sum would be 2+1+... but the answer choices are 8,9,11, etc. So maybe I messed up.Wait, perhaps my approach is wrong. Let me check for r=2 again. If log10(N)≈312.8908555306, then log10(N^(1/2))≈156.4454277653. The fractional part is 0.4454277653.10^{0.4454277653} ≈ ?log10(2.78) ≈ 0.444, log10(2.79)≈0.4456. So 0.445427 is just a hair below 0.4456, so 10^{0.445427}≈2.79? Wait, wait.Wait, let me cross-check. If log10(2.79)≈0.4456, then 10^{0.445427}≈2.79 - a little bit. Wait, but 0.445427 is 0.4456 - 0.000173. So the difference between 0.445427 and 0.4456 is 0.000173. Since log10(2.79) = 0.4456, so 10^{0.4456} =2.79. Therefore, 10^{0.4456 -0.000173} = 2.79 / 10^{0.000173} ≈ 2.79 / (1 + 0.000173 * ln10) ≈ 2.79 / (1 + 0.0003986) ≈ 2.79 / 1.0003986 ≈ 2.789. So approximately 2.789 - but wait, no, 2.79 divided by approximately 1.0004 is approximately 2.788. So 10^{0.445427} ≈2.788. Therefore, leading digit 2. Therefore, f(2)=2.Wait, but then if f(2)=2, f(3)=1, then let's proceed to r=4.For r=4:log10(N^(1/4)) ≈ 312.8908555306 /4 ≈78.22271388265So integer part 78, fractional part 0.22271388265Compute 10^{0.22271388265}10^{0.2227} ≈ ?We know that log10(1.67) ≈0.2227. Let me check: log10(1.67) ≈0.2227. So 10^{0.2227} ≈1.67. Therefore, leading digit is 1. Wait, 1.67 starts with 1. So f(4)=1?Wait, but 1.67 is the number, so leading digit is 1. But let me verify with precise calculation.Again, let's compute 10^{0.22271388265} = e^{0.22271388265 * ln10} ≈ e^{0.22271388265 * 2.302585093} ≈ e^{0.51273}Compute e^{0.51273}. e^0.5 ≈1.64872, e^0.51273 is a bit higher.Compute e^{0.51273} = e^{0.5 +0.01273} ≈ e^0.5 * e^{0.01273} ≈1.64872 * (1 +0.01273 +0.01273^2/2 + ...) ≈1.64872*(1.0128 +0.00008)≈1.64872*1.01288≈1.64872 +1.64872*0.01≈1.64872 +0.016487≈1.6652. So approximately 1.6652. So 1.6652, leading digit 1. Therefore, f(4)=1.Wait, but 1.6652 is 1.66..., leading digit 1. So f(4)=1.Hmm, so now f(2)=2, f(3)=1, f(4)=1. Then r=5:log10(N^(1/5))≈312.8908555306 /5≈62.57817110612Integer part 62, fractional part 0.57817110612Compute 10^{0.57817110612}10^{0.5781711} ≈ ?We know that log10(3.75)≈0.5740, log10(3.8)=0.5798.So 0.5781711 is between log10(3.75) and log10(3.8). Let's see:log10(3.75)=0.574031log10(3.8)=0.5797840.5781711 is 0.5781711-0.574031=0.0041401 above 0.574031.The interval between 3.75 and 3.8 is 0.05 in linear scale, and the log difference is 0.579784 -0.574031=0.005753.So 0.0041401 /0.005753≈0.719 of the interval. Therefore, 3.75 +0.719*0.05≈3.75+0.03595≈3.78595. So approximately 3.786. Therefore, 10^{0.5781711}≈3.786. Therefore, leading digit is 3. So f(5)=3.Alternatively, compute 10^{0.5781711}:Take ln(10^{0.5781711})=0.5781711*ln(10)≈0.5781711*2.302585≈1.330e^{1.330} ≈ e^{1.3}*e^{0.03} ≈3.6693*1.03045≈3.6693*1.03≈3.78. So 3.78, leading digit 3. Therefore, f(5)=3.So f(5)=3.Proceeding to r=6:log10(N^(1/6))≈312.8908555306 /6≈52.14847592177Integer part 52, fractional part 0.14847592177Compute 10^{0.14847592177}10^{0.14847592177}≈?We know that log10(1.4)=0.1461, log10(1.41)=0.1491.0.148475 is between those. Let's compute.log10(1.4)=0.1461log10(1.41)=0.1491Difference in log: 0.1491 -0.1461=0.003The value 0.148475 is 0.148475 -0.1461=0.002375 above log10(1.4). So 0.002375 /0.003≈0.7917 of the interval from 1.4 to 1.41.Therefore, 1.4 +0.7917*0.01≈1.4 +0.007917≈1.4079. So approximately 1.408. Therefore, 10^{0.148475}≈1.408. Leading digit is 1. Therefore, f(6)=1.But let me verify with exponent:10^{0.14847592177}= e^{0.14847592177*ln10}≈e^{0.14847592177*2.302585}≈e^{0.3417}Compute e^{0.3417}≈e^{0.3}≈1.349858, e^{0.3417}= e^{0.3+0.0417}=1.349858 * e^{0.0417}≈1.349858 *1.0425≈1.349858*1.04≈1.403 and 1.349858*0.0025≈0.00337, so total≈1.40637. So approximately 1.406. Leading digit 1. Therefore, f(6)=1.Therefore, summing up:f(2)=2f(3)=1f(4)=1f(5)=3f(6)=1Total sum: 2+1+1+3+1=8. But 8 is option A. However, the answer choices are (A) 8, (B) 9, (C)11, (D)22, (E)29.But the answer isn't 8? Wait, but according to my calculations, the sum is 8. However, the options include 8 as option A. So maybe that's the answer? But wait, the answer is supposed to be among these. Wait, but let me check my steps again because maybe I made an error.Wait, let me verify each f(r):For r=2: calculated leading digit 2. Is that correct? Let me check again.log10(N^(1/2))≈156.4454277653, fractional part≈0.445427765310^{0.4454277653}≈2.787, leading digit 2. Correct.r=3:≈1.98, leading digit 1. Correct.r=4:≈1.665, leading digit 1. Correct.r=5:≈3.78, leading digit 3. Correct.r=6:≈1.406, leading digit 1. Correct.Sum:2+1+1+3+1=8. So answer is A)8.But according to the original problem statement, the answer choices are (A) 8, (B) 9, (C)11, (D)22, (E)29.But wait, perhaps I made a miscalculation in one of the roots. Let me check each step again.First, verify log10(N)=312.8908555306.Yes, since N is 7 followed by 312 7s, which is 7*(10^313 -1)/9. So log10(N)=log10(7) + log10(10^313 -1) - log10(9). Since log10(10^313 -1)≈313, so log10(N)=log10(7) - log10(9) +313≈313 -0.109144≈312.890856. Correct.Then, for r=2:312.890856 /2=156.445428. Fractional part 0.445428.10^{0.445428}= approx 2.787. Leading digit 2. Correct.r=3:312.890856 /3=104.296952. Fractional part 0.296952. 10^{0.296952}=1.9804. Leading digit 1. Correct.r=4:312.890856 /4=78.222714. Fractional part 0.222714. 10^{0.222714}=1.67. Leading digit 1. Correct.r=5:312.890856 /5=62.578171. Fractional part 0.578171. 10^{0.578171}=3.78. Leading digit 3. Correct.r=6:312.890856 /6=52.148476. Fractional part 0.148476. 10^{0.148476}=1.406. Leading digit 1. Correct.Sum=2+1+1+3+1=8. Answer A.But the options given include A:8. So unless there is a miscalculation here, the answer should be 8. However, the original problem is from an Olympiad, and answer choices often don't include the first option unless it's a trick question. But maybe I made a mistake in approximating the logs? Let me check for r=5 again, because 3.78 is close to 3.8, which would start with 3, but maybe the actual value is higher?Wait, let's redo the calculation for r=5.log10(N^(1/5))=62.57817110612. The fractional part is 0.57817110612.Compute 10^{0.57817110612}. Let's use more precise method.0.57817110612*ln10=0.57817110612*2.302585093≈0.57817110612*2.302585093.Compute 0.5*2.302585=1.15129250.07817110612*2.302585093≈0.0781711*2.302585≈0.1801Total≈1.1512925+0.1801≈1.3313925Therefore, e^{1.3313925}≈e^{1.33}≈3.785. Let's compute e^1.3313925.We can note that e^1.331 ≈3.78. Specifically:e^1 = 2.71828e^0.33 ≈1.39097Therefore, e^1.33≈2.71828*1.39097≈2.71828*1.39≈3.778Therefore, e^{1.3313925}≈3.78. So leading digit 3. Correct.Therefore, f(5)=3.Thus, sum is 8. Answer A.But the problem is from an Olympiad, and answer A is 8. But I need to check again if I made any miscalculations.Alternatively, perhaps the question is about the leading digit of the rth root, not the integer part. Wait, but the rth root of N is a number that when raised to the rth power gives N. Since N is an integer, the rth root is not necessarily an integer, but we just need the leading digit of the real number result. So, for example, the square root of N is a number starting with 2.78..., so leading digit 2. Similarly, cube root starts with 1.98..., leading digit 1. All correct.Wait, but maybe the problem defines the rth root as the integer part? For example, the greatest integer less than or equal to the root. But the problem says "the leading digit of the rth root of N". So, if the root is 1.98..., then leading digit is 1. If it's 3.78..., leading digit 3. So my calculation is correct.Alternatively, maybe there is a mistake in the problem's answer choices. But original problem statement says the options are (A)8, (B)9, (C)11, (D)22, (E)29. So according to my calculation, the answer is A)8. But I need to check again.Wait, let me check for r=6. log10(N^(1/6))≈52.14847592177, fractional part≈0.1484759217710^{0.14847592177}≈1.406. Leading digit 1. Correct.Alternatively, maybe using more accurate log10(N). Suppose we calculate log10(N) more precisely. Let's compute log10(N)=log10(7)+log10((10^313 -1)/9). Since (10^313 -1)/9 is a number with 313 1s. So log10((10^313 -1)/9)=log10(10^313 -1) - log10(9)=313 - ε - log10(9). So log10(N)=log10(7)+313 - ε - log10(9)=313 + log10(7) - log10(9) - ε. Since ε is negligible (≈10^{-313}/ln10), so log10(N)=312.8908555306 - ε. Therefore, log10(N) is slightly less than 312.8908555306.Therefore, when calculating for r=3, log10(N)/3≈104.2969518435 - ε/3. Therefore, the fractional part is 0.2969518435 - ε/3. But since ε is so small, the fractional part is still approximately 0.2969518435. So 10^{0.2969518435}≈1.9804, leading digit 1. Correct.Therefore, all steps are correct, leading to sum=8. Answer A. But according to the original problem's answer choices, A is 8. So that must be the correct answer. However, in the original problem statement, it's possible that the test-maker expects the answer to be different. Wait, but how?Wait, another thought: Maybe I messed up the number of digits. The problem states N is a 313-digit number, each digit 7. So N has 313 digits, all 7s. So N=777...777 with 313 digits. Therefore, log10(N) is between log10(7*10^{312}) and log10(8*10^{312}), which is log10(7)+312 ≈0.8451 +312=312.8451 and log10(8)+312≈0.9031 +312=312.9031. But when I calculated log10(N) as≈312.890855, which is within this interval. But wait, if all digits are 7, N is much closer to 7.777...*10^{312}, so log10(N)=log10(7.777...*10^{312})=log10(7.777...) +312. What's log10(7.777...)?Since 7.777... is 7 + 0.7 +0.07 +... up to 313 terms. But it's finite. So 7.777...777 with 313 digits. Wait, but this is equal to 7*(10^{313}-1)/9*10^{-0}? Wait, no. Wait, N is 777...777 with 313 digits. So it's 7*10^{312} +7*10^{311} +...+7*10^0. Which is equal to 7*(10^{313}-1)/9. As before. Therefore, log10(N)=log10(7) + log10(10^{313}-1) - log10(9). As calculated before≈312.890855. Therefore, correct.Therefore, my calculation is correct. The answer should be 8, option A. But let me check once more for r=5. Perhaps I miscalculated.log10(N^(1/5))≈62.57817110612, fractional part≈0.57817110612.10^{0.57817110612}=approx 3.78, leading digit 3. Correct.Alternatively, maybe I used the wrong fractional part. Let me check:For example, log10(N^(1/5))≈62.57817110612. So the fractional part is 0.57817110612. So 10^{0.5781711}≈3.78. Correct.Another way: Take 10^{0.5781711}=10^{0.57}*10^{0.0081711}. 10^{0.57}=approx 3.715, 10^{0.0081711}=approx1.019. So 3.715*1.019≈3.785. Leading digit 3. Correct.Therefore, all steps correct. Therefore, the answer is 8. So option A.But I recall that in some problems similar to this, the answer might not be A. Maybe the test-maker made a mistake? Or maybe I missed something. Wait, but according to the exact calculation, answer is 8. Maybe in the original problem, the answer is different. Wait, let me check the original problem statement again.Problem statement:Let N be the positive integer 7777…777, a 313-digit number where each digit is a 7. Let f(r) be the leading digit of the rth root of N. What is f(2) + f(3) + f(4) + f(5)+ f(6)? Answer choices: (A) 8, (B)9, (C)11, (D)22, (E)29.No, seems my answer is correct. Maybe the answer is 8. But in the original problem, maybe options are different. Wait, no, in the problem statement, the options are given as (A) 8, (B) 9, (C) 11, (D) 22, (E) 29. So 8 is an option, and my calculation gives 8. Therefore, answer is A.But I need to check again if the leading digit of the cube root is indeed 1. Let me check with an example. Suppose N is 777 (3-digit number). Then N=777. The cube root of 777 is approx 9.19, leading digit 9. But wait, that contradicts our previous method. Wait, what's happening here.Wait, take N=777 (3 digits). log10(N)=log10(777)=2.8903. Then cube root: log10(N)/3≈0.9634. 10^{0.9634}≈9.19, leading digit 9. So f(3)=9. But according to our previous method, for a 3-digit N=777, log10(N)=log10(7*100 +7*10 +7)=log10(777)≈2.8903. So 2.8903/3≈0.9634, 10^{0.9634}≈9.19. Leading digit 9. Therefore, in this case, f(3)=9.Wait, but in our original problem, for a 313-digit N, the cube root leading digit was 1. So why the difference? Because for the 3-digit number, log10(N)/3 is less than 1, so the leading digit is in the units place. But in the case of the 313-digit number, log10(N)/3 is 104.296..., so the leading digit is in the 10^104 scale, and the fractional part 0.29695 gives 10^{0.29695}≈1.98, leading digit 1. So the leading digit depends on the fractional part of log10(N)/r.But in the 3-digit case, the cube root is a single-digit number (9.19), but in the 313-digit case, the cube root is a 105-digit number starting with 1.98...×10^104. Therefore, leading digit 1.So in the case of the 3-digit number, the cube root is a 1-digit number with leading digit 9. In the 313-digit number, the cube root is a 105-digit number starting with 1. Therefore, leading digit 1.Therefore, my calculation is correct. Therefore, the answer is 8. So the answer should be A)8.But according to the problem's answer choices, 8 is an option. Therefore, I think the correct answer is A)8.Final Answerboxed{A}</think>Given ( N ), a 313-digit number where each digit is 7, we need to find the leading digit of the ( r )-th root of ( N ) for ( r = 2, 3, 4, 5, 6 ) and sum these leading digits.1. Calculate ( log_{10}(N) ): - ( N ) is a 313-digit number with all digits 7, which can be approximated as ( 7 times 10^{312} ). - Using the exact expression ( N = frac{7}{9}(10^{313} - 1) ), we approximate ( log_{10}(N) approx log_{10}(7) - log_{10}(9) + 313 ). - ( log_{10}(7) approx 0.845098 ), ( log_{10}(9) approx 0.954243 ). - ( log_{10}(N) approx 313 + 0.845098 - 0.954243 = 312.890855 ).2. Calculate leading digits for each ( r ): - For each ( r ), compute ( frac{log_{10}(N)}{r} ), extract the fractional part, and compute ( 10^{text{fractional part}} ). - ( r = 2 ): - ( frac{312.890855}{2} = 156.445427 ) - Fractional part: ( 0.445427 ) - ( 10^{0.445427} approx 2.78 ), leading digit: 2. - ( r = 3 ): - ( frac{312.890855}{3} = 104.296952 ) - Fractional part: ( 0.296952 ) - ( 10^{0.296952} approx 1.98 ), leading digit: 1. - ( r = 4 ): - ( frac{312.890855}{4} = 78.222714 ) - Fractional part: ( 0.222714 ) - ( 10^{0.222714} approx 1.67 ), leading digit: 1. - ( r = 5 ): - ( frac{312.890855}{5} = 62.578171 ) - Fractional part: ( 0.578171 ) - ( 10^{0.578171} approx 3.78 ), leading digit: 3. - ( r = 6 ): - ( frac{312.890855}{6} = 52.148476 ) - Fractional part: ( 0.148476 ) - ( 10^{0.148476} approx 1.41 ), leading digit: 1.3. Sum the leading digits: - ( f(2) + f(3) + f(4) + f(5) + f(6) = 2 + 1 + 1 + 3 + 1 = 8 ).Thus, the final answer is (boxed{A}).]

🔑:We can rewrite N as frac{7}{9}cdot 9999ldots999 = frac{7}{9}cdot(10^{313}-1). When approximating values, as we will shortly do, the minus one will become negligible so we can ignore it.When we take the power of ten out of the square root, we’ll be multiplying by another power of ten, so the leading digit will not change. Thus the leading digit of f(r) will be equal to the leading digit of sqrt[r]{frac{7}{9} cdot 10^{313(mod r)}}.Then f(2) is the first digit of sqrt{frac{7}{9}cdot(10)} = sqrt{frac{70}{9}} = sqrt{7.ldots} approx 2f(3) - sqrt[3]{frac{7}{9} cdot 10} = sqrt[3]{frac{70}{9}} = sqrt[3]{7.ldots} approx 1.f(4) - sqrt[4]{frac{7}{9} cdot 10} = sqrt[4]{frac{70}{9}} = sqrt[4]{7.ldots} approx 1.f(5) - sqrt[5]{frac{7}{9} cdot 1000} = sqrt[5]{frac{7000}{9}} = sqrt[5]{777.ldots} approx 3.f(6) - sqrt[6]{frac{7}{9} cdot 10} = sqrt[6]{frac{70}{9}} = sqrt[6]{7.ldots} approx 1.The final answer is therefore 2+1+1+3+1 = boxed{textbf{(A) }8}.~KingRavi

🧠:[<think>Okay, so I need to solve this geometry problem about a zig-zag line on a circle. Let me try to visualize it first. There's a circle with diameter PQ. The zig-zag line starts at point P, which is one end of the diameter, and after four peaks, it ends at point Q. Each angle between the zig-zag line and the diameter PQ is α. The question is asking for the measure of angle α, and the options are given from 60° up to 86°. Hmm.First, let me sketch this mentally. The circle has diameter PQ. So P and Q are at opposite ends. The zig-zag starts at P, goes towards the circumference, makes a peak, then back towards the diameter, but not necessarily at the center each time. Each time it meets the diameter, it creates an angle α. After four such peaks, it ends at Q. So each "zig" or "zag" is a segment that makes an angle α with the diameter.Wait, the problem says "each of the angles between the zig-zag line and the diameter PQ is equal to α." So every time the zig-zag meets the diameter PQ, the angle it makes with PQ is α. Since it's a zig-zag, it alternates direction each time. So after starting at P, it goes up making angle α with PQ, then hits the circumference, reflects off at angle α on the other side, and so on. After four peaks, it reaches Q.But how does this work exactly? Each time the line hits the circumference, it reflects such that the angle with the diameter is α again. So it's similar to the law of reflection, where the angle of incidence equals the angle of reflection. But instead of reflecting off a surface, it's reflecting off the circumference? Wait, no. The diameter is PQ, and the zig-zag line is bouncing off the diameter each time, making angle α with PQ. Wait, but the diameter is a straight line through the center. If the zig-zag is bouncing between the top and bottom of the diameter? Wait, maybe I need to clarify.Alternatively, perhaps the zig-zag is going from P, making a peak above the diameter, then crossing the diameter at some point, making another peak below, crossing again, and so on. Each time it crosses the diameter, the angle between the zig-zag line and PQ is α. So there are four peaks, meaning it crosses the diameter four times between P and Q? Wait, but starting at P, the first segment goes from P, makes a peak, crosses the diameter at some point, then makes another peak, crosses again, etc., and after four peaks, ends at Q.Wait, if there are four peaks, then how many times does it cross the diameter? Let's see. Starting at P, the first segment goes up (or down) making angle α with PQ, then hits the circumference. Then it reflects, making angle α again with PQ, goes back down (or up), crossing the diameter, then continues. Each peak is a reflection off the circumference. Wait, but the circumference is a circle, so the reflection would depend on the tangent at that point.Alternatively, maybe each "peak" is a point where the zig-zag changes direction, either at the circumference or crossing the diameter? The problem is a bit unclear, but I need to parse it correctly.Wait, the problem says: "A zig-zag line starts at the point P, at one end of the diameter PQ of a circle. Each of the angles between the zig-zag line and the diameter PQ is equal to α as shown. After four peaks, the zig-zag line ends at the point Q." So the key points are:- Starts at P (end of diameter PQ)- Each angle between the zig-zag and PQ is α- Four peaks, then ends at QSince there's a diagram shown (but I can't see it), the peaks are likely the points where the zig-zag changes direction on the circumference. So each time it reaches the circumference, it "peaks" there, then heads back towards the diameter, making angle α with PQ. So starting at P, goes towards the circumference, makes a peak, then goes back towards the diameter, crossing it at some point, then again towards the circumference, peaks again, and so on, until after four peaks, it ends at Q.So each "peak" is a point on the circumference where the zig-zag turns. Each time it approaches the diameter, it crosses it at some point, making angle α with PQ. So between each peak, the zig-zag crosses the diameter once. Therefore, four peaks would mean it crosses the diameter three times between P and Q. Wait, starting at P, then first segment to first peak (on circumference), then back down to first crossing point on diameter, then up to second peak, then back down to second crossing point, then up to third peak, back to third crossing, then up to fourth peak, then back down to Q. So from P to peak1, peak1 to cross1, cross1 to peak2, peak2 to cross2, cross2 to peak3, peak3 to cross3, cross3 to peak4, peak4 to Q. So total segments: P-peak1, peak1-cross1, cross1-peak2, peak2-cross2, cross2-peak3, peak3-cross3, cross3-peak4, peak4-Q. That's 8 segments. But peaks are the points on the circumference: peak1, peak2, peak3, peak4. Then the crossings are on the diameter: cross1, cross2, cross3. So after four peaks, starting at P, crossing the diameter three times, then ending at Q.Alternatively, maybe the "peaks" refer to the number of times it turns direction on the circumference. So each time it goes from diameter to circumference (a peak) and back. So four peaks would mean four times going from diameter to circumference and back. But that would be two crossings? Wait, perhaps the number of crossings on the diameter is equal to the number of peaks? Wait, no.Wait, perhaps it's helpful to think of this as a polygonal path that starts at P, makes four "peaks" (turning points on the circumference), and ends at Q. Each time it approaches the diameter, it makes an angle α with PQ. So each segment of the zig-zag makes an angle α with the diameter PQ.Given that it's a circle, the reflections off the circumference would follow the law of reflection, where the angle of incidence equals the angle of reflection with respect to the tangent at the point of incidence. But in this problem, it's given that the angles between the zig-zag and the diameter PQ are all equal to α. So perhaps regardless of the point where the zig-zag meets the diameter, the angle between the zig-zag segment and PQ is always α. That might not be the usual reflection, unless the diameter is acting as some axis.Alternatively, maybe the path is constructed such that each time it meets the diameter PQ, the angle between the incoming segment and PQ is α, and the same for the outgoing segment. But if it's bouncing off the diameter, then the angle of incidence would equal the angle of reflection. So if it comes in at angle α below the diameter, it reflects at angle α above, but since it's a diameter, which is a straight line, the reflection over the diameter would be symmetric. Wait, but the problem says "each of the angles between the zig-zag line and the diameter PQ is equal to α". So maybe all the angles, both above and below, are α. Wait, but if you have a zig-zag crossing the diameter each time, then the angle on one side would be α above PQ, and α below PQ. But since PQ is a straight line, these angles would be congruent. So each time the zig-zag meets PQ, it forms an angle α with PQ, alternating sides.But how does this lead to a specific value for α? The key is that after four such peaks (i.e., four turns), the path starts at P and ends at Q. So the total "deviation" caused by each angle α accumulates over the four segments to direct the path from P to Q.Wait, perhaps this is similar to light reflecting multiple times between two mirrors, but in this case, the "mirror" is the diameter PQ, and the path reflects off the circumference each time. Wait, maybe not. Alternatively, since the path is constrained to the circle, each peak is a point on the circumference where the path turns by angle 2α? Hmm, maybe.Alternatively, think of the circle with diameter PQ. The path starts at P, goes in some direction making angle α with PQ. This direction will intersect the circumference at some point, say A1. Then at A1, the path turns by 180° - 2α (since the angle with PQ is again α on the other side). Then from A1, it goes to intersect PQ at some point B1, making angle α with PQ. Then from B1, it goes to another point A2 on the circumference, making angle α, and so on, until after four peaks (A1, A2, A3, A4), it reaches Q.Each time it goes from a point on PQ to the circumference, making angle α, reflects (changes direction by 180° - 2α), goes back to PQ, and so on. So each "bounce" off the circumference changes the direction by 180° - 2α. After four such bounces, the total rotation would result in the direction from the last point to Q.Alternatively, each segment from PQ to circumference is a chord of the circle making angle α with PQ. The cumulative effect after four such segments spans the diameter PQ. So perhaps the sum of the projections of each chord onto the diameter PQ equals the length of PQ.Let me formalize this.Let’s denote the circle with diameter PQ, so the length of PQ is 2R, where R is the radius. The path starts at P, goes to a point A1 on the circumference, making angle α with PQ. Then from A1, it goes to a point B1 on PQ, making angle α with PQ (on the other side), then to A2, etc., until after four peaks (four times reaching the circumference), it ends at Q.Each segment from PQ to the circumference is a chord. Let's consider the horizontal component (along PQ) of each chord. Since the angle with PQ is α, the horizontal component of each chord is 2R cos α, but wait, the chord length is 2R sin θ, where θ is the angle between the radius and the chord. Wait, maybe I need to compute the actual horizontal component.Wait, if a chord makes an angle α with PQ (the diameter), then the horizontal component (along PQ) of the chord would be the chord length multiplied by cos α. But the chord length can be related to the central angle.Alternatively, consider the chord from P to A1. The angle between chord PA1 and PQ is α. Since PQ is the diameter, the chord PA1 makes angle α with PQ. The length of PA1 can be found using trigonometry. Let’s denote the central angle corresponding to chord PA1 as 2θ. Then θ is the angle between the radius OP and PQ, where O is the center. Wait, if chord PA1 makes angle α with PQ, then the angle between PA1 and the radius at A1 is 90° - α, because the radius is perpendicular to the tangent at A1, and the chord PA1 makes angle α with PQ. Hmm, maybe this is getting complicated.Alternatively, let's model the coordinate system. Let’s place the circle with diameter PQ along the x-axis, with center at the origin O. So P is at (-R, 0), Q is at (R, 0). The zig-zag path starts at P and makes an angle α above the x-axis. This first segment will intersect the circumference at some point A1. The equation of this line is y = tan α (x + R). The circle equation is x² + y² = R². Substituting y = tan α (x + R) into the circle equation:x² + [tan² α (x + R)^2] = R²x² + tan² α (x^2 + 2R x + R²) = R²x² (1 + tan² α) + 2 R tan² α x + R² tan² α - R² = 0Using the identity 1 + tan² α = sec² α:x² sec² α + 2 R tan² α x + R² (tan² α - 1) = 0Multiply through by cos² α to eliminate denominators:x² + 2 R sin² α x + R² (sin² α - cos² α) = 0This quadratic equation in x can be solved. However, since we know that x = -R is a root (since the line passes through P(-R, 0)), we can factor it out.Let me check: The line from P(-R, 0) with angle α will intersect the circle again at A1. So factoring (x + R):(x + R)(x - x1) = 0Expanding:x² + ( - x1 + R )x - R x1 = 0Comparing with the previous equation:x² + 2 R sin² α x + R² (sin² α - cos² α) = 0So coefficients:- x1 + R = 2 R sin² α => -x1 = 2 R sin² α - R => x1 = R - 2 R sin² α = R(1 - 2 sin² α) = R cos 2αAnd:- R x1 = R² (sin² α - cos² α) => -x1 = R (sin² α - cos² α) => x1 = -R (sin² α - cos² α) = R (cos² α - sin² α) = R cos 2αConsistent. So the x-coordinate of A1 is R cos 2α. Then the y-coordinate is tan α (x + R) = tan α (R cos 2α + R) = tan α * R (cos 2α + 1). Using the identity cos 2α + 1 = 2 cos² α, this becomes tan α * R * 2 cos² α = 2 R sin α cos α = R sin 2α. So A1 is at (R cos 2α, R sin 2α).So from P(-R, 0), the first segment goes to A1(R cos 2α, R sin 2α). Then from A1, the path goes to the next point B1 on PQ, making angle α with PQ, but on the other side. Since it's a zig-zag, the angle is now below the x-axis, i.e., with angle -α. So the line from A1 has slope -tan α.Let me verify: The angle with PQ is α, but since we're going from A1 to B1, which is on PQ. If the angle is measured from PQ, then if we are below the x-axis, the slope is -tan α.The equation of the line from A1(R cos 2α, R sin 2α) with slope -tan α is y - R sin 2α = -tan α (x - R cos 2α). To find where it intersects PQ (the x-axis), set y = 0:0 - R sin 2α = -tan α (x - R cos 2α)=> - R sin 2α = -tan α x + R tan α cos 2αMultiply both sides by -1:R sin 2α = tan α x - R tan α cos 2αThen,tan α x = R sin 2α + R tan α cos 2αDivide both sides by tan α:x = R [ sin 2α / tan α + cos 2α ]Note that sin 2α / tan α = sin 2α / (sin α / cos α) ) = sin 2α * cos α / sin α = 2 sin α cos α * cos α / sin α = 2 cos² αTherefore,x = R [ 2 cos² α + cos 2α ]But cos 2α = 2 cos² α - 1, so:x = R [ 2 cos² α + 2 cos² α - 1 ] = R [4 cos² α - 1]So point B1 is at (R(4 cos² α - 1), 0)Now, this is the x-coordinate where the second segment intersects PQ. So starting from P(-R, 0), after one peak (A1), the path intersects PQ at B1(R(4 cos² α - 1), 0). Then from B1, the path goes to A2, making angle α above the x-axis again?Wait, no. Wait, the problem says "each of the angles between the zig-zag line and the diameter PQ is equal to α". So regardless of the direction, whether going from P to A1 or from A1 to B1, the angle with PQ is α. But when going from A1 to B1, the angle is below the x-axis, so the angle is still α, but on the other side.Therefore, each time the path crosses PQ, it alternates direction, making angle α above and below. So the process repeats: from B1, the path goes to A2 on the circumference, making angle α above PQ, then to B2 on PQ making angle α below, etc.So let's try to find the pattern here. Starting at P(-R, 0):1. First segment: P to A1(R cos 2α, R sin 2α)2. Second segment: A1 to B1(R(4 cos² α - 1), 0)3. Third segment: B1 to A2, which would be similar to the first segment but starting from B1. Let's compute A2.From B1(R(4 cos² α - 1), 0), the path makes an angle α above PQ. The equation of this line is y = tan α (x - x_B1). Let's compute x_B1 = R(4 cos² α - 1). So the equation is y = tan α (x - R(4 cos² α - 1)).To find the intersection A2 with the circle, substitute into x² + y² = R²:x² + [tan² α (x - R(4 cos² α - 1))^2] = R²Similar to before, this will result in a quadratic equation. Since we know that x = x_B1 is a root (the line passes through B1), the other root will give the x-coordinate of A2.Alternatively, notice that the pattern from P to A1 to B1 involved a reflection or a symmetry. Maybe there's a recursive formula here.Alternatively, perhaps each intersection with PQ is a linear transformation. Let's see.From P(-R, 0), after the first segment, we reach B1 at x = R(4 cos² α - 1). Then from B1, the next intersection with PQ after another peak would be applying the same transformation. So starting from x_n, the next x_{n+1} is given by x_{n+1} = R(4 cos² α - 1) + (x_n - R(4 cos² α - 1)) * something?Wait, perhaps not. Let me think differently. The horizontal distance covered from P to B1 is R(4 cos² α - 1) - (-R) = R(4 cos² α - 1 + 1) = 4 R cos² α. So the first segment (from P to A1 to B1) covers a horizontal distance of 4 R cos² α.Similarly, each subsequent peak would cover a horizontal distance multiplied by 2 cos 2α? Wait, maybe not. Wait, in the first step, moving from P(-R, 0) to B1(R(4 cos² α - 1), 0), the total horizontal progress is 4 R cos² α. If this is a geometric progression, then each subsequent bounce covers a fraction of the remaining distance. But after four peaks, the total horizontal distance covered should be 2R (from -R to R).Wait, the total horizontal distance from P to Q is 2R. Each "peak" corresponds to a horizontal component. If each segment pair (up and down) contributes a certain horizontal distance, then after four peaks, the sum of these contributions equals 2R.Wait, maybe each "zig" (from PQ to circumference and back) contributes a horizontal distance. Let's consider that. The first zig (P to A1 to B1) covers 4 R cos² α. Then the next zig (B1 to A2 to B2) would cover the same horizontal distance, but scaled by some factor. Wait, but in the first step, moving from x = -R to x = R(4 cos² α - 1). The distance is 4 R cos² α. Then moving from B1 to B2 would be another 4 R cos² α times (something). Wait, if each zig-zag reduces the remaining distance by a factor, perhaps?Alternatively, perhaps the horizontal progression each time is multiplied by 2 cos 2α. Because when you go from P to A1 to B1, the horizontal component is 4 R cos² α. Then from B1, the next horizontal component would be 4 R cos² α multiplied by 2 cos 2α? Not sure. Let me check with angles.Wait, when you go from P to A1, the angle is α, so the horizontal component is 2R cos α (the chord length is 2R sin θ, but the horizontal component would be chord length * cos α). Wait, maybe my initial approach was better with coordinates.From the first segment, we found that the horizontal progress from P to B1 is 4 R cos² α. If we can model the total distance after four peaks as the sum of four such contributions, each time with a different multiplier. But maybe each subsequent horizontal component is multiplied by a factor of 2 cos 2α. Let's see.Wait, actually, after the first horizontal progress of 4 R cos² α, the remaining distance to Q is 2R - 4 R cos² α. Then the next horizontal progress would be something else. Alternatively, maybe each peak corresponds to a reflection and there's a geometric sequence here.Alternatively, think of the problem as analogous to light reflecting multiple times between two surfaces, but in this case, the path is constrained by the circle. Each "reflection" at the circumference redirects the path by an angle dependent on α, and after four reflections, the total deviation results in the path ending at Q.But perhaps a better approach is to use the theory of billiards. In a circular billiard table, the path of a ball reflects off the boundary with equal angles. However, in this problem, the reflection is such that the angle with the diameter PQ is always α. So maybe unfolding the path.In billiards problems, unfolding the table (reflecting the table across the cushion) is a common technique to turn a reflected path into a straight line. Similarly, here, we can "unfold" the circle each time the path reflects, creating a straight line path that goes through multiple copies of the circle.Since each reflection over the diameter PQ flips the direction across PQ, unfolding the circle across PQ would place a mirrored circle. Then the zig-zag path in the original circle becomes a straight line in the unfolded plane passing through multiple circles. The total number of reflections (peaks) corresponds to the number of circles we pass through.In this case, after four peaks (four reflections), the path ends at Q, which is the antipodal point of P. So unfolding the path four times would create a straight line from P to the fourth copy of Q in the unfolded plane.But perhaps more precise: each reflection over PQ can be represented by mirroring the circle across PQ. So starting at P, the first segment goes to A1 in the original circle. Upon reflection, instead of changing direction, we can mirror the circle across PQ, and the path continues straight into the mirrored circle. So after four reflections, we would have four mirrored circles, and the straight line path from P in the original circle to Q in the fourth mirrored circle.The angle α then corresponds to the angle between this straight line and PQ. The total distance traveled in the unfolded plane would be the length of the straight line from P to Q4 (the fourth mirrored Q), and the horizontal component (along PQ) would be the sum of the diameters of the circles, which is 2R * 4 = 8R, but the vertical component remains the same, but since we end at Q, which is on the diameter, the vertical components must cancel out.Wait, perhaps this is confusing. Let me formalize.In billiards unfolding, each reflection over a line can be represented by reflecting the plane across that line and continuing the straight path. So for each reflection over PQ, we mirror the plane, resulting in a straight path through multiple mirrored planes.If the original circle is reflected over PQ four times, we get four copies of the circle in a straight line along PQ. The path from P to Q with four reflections would correspond to a straight line from P in the original circle to Q in the fourth mirrored circle.The angle α in the original problem corresponds to the angle between this straight line and the line PQ in the unfolded plane.So the total horizontal distance in the unfolded plane is 4 * 2R = 8R (since each reflection adds a diameter length). The vertical distance must be zero because the final point Q is on the diameter PQ. Therefore, the straight line in the unfolded plane goes from P to Q4, which is 8R away horizontally. But since the vertical component is zero, the angle α must be such that the vertical components accumulated over each segment cancel out.Wait, perhaps not. Wait, in the unfolded plane, the path is a straight line from P to Q4, which is 8R to the right. However, in reality, the path bounces up and down, but in the unfolded plane, it's a straight line with slope determined by the vertical deviation.Wait, but if after four reflections, the total vertical displacement must be zero (since we end at Q on PQ), then the total "up" and "down" movements must cancel out. Therefore, in the unfolded plane, the straight line path must have a slope such that over 8R horizontal distance, the vertical displacement is zero. Therefore, the angle α is zero, which contradicts. Hmm, maybe this approach isn't correct.Wait, no. Because in reality, each reflection alternates the direction of the vertical component. So in the unfolded plane, each mirrored circle alternates the vertical direction. Wait, no. If we unfold over PQ, which is a horizontal line, then each reflection over PQ would invert the vertical coordinate. So the mirrored circles above and below PQ. But since all reflections are over PQ (the diameter), each reflection flips the circle vertically. So after four reflections, we would have two circles above and two below, but the final point Q is in the fourth reflected circle, which is on the same horizontal line as the original.Wait, maybe the path in the unfolded plane is a straight line from P to the fourth reflection of Q, which is 8R to the right, but in alternating up/down positions. However, since we end at Q, which is 2R away from P, this seems conflicting.Alternatively, perhaps the number of reflections is different. Wait, the problem says "after four peaks", which are the points where the path touches the circumference. Each peak corresponds to a reflection, so four peaks mean four reflections. In billiards unfolding, four reflections would correspond to four mirrored circles. The straight line path would go through four circles, with the angle α determining the slope.The straight line distance in the unfolded plane would be the hypotenuse of a right triangle with horizontal leg 4*(2R) = 8R and vertical leg zero, but that would require α = 0, which isn't the case. So maybe my analogy is wrong.Alternatively, if each peak corresponds to a chord across the circle making angle α with PQ. Then after four such chords, the total horizontal displacement sums up to 2R.Each chord has a horizontal component of 2R cos α (the projection onto PQ). But since the path zig-zags, the horizontal components alternate direction? Wait, no. Each chord goes from a point on PQ to another point on PQ, but displaced by 2R cos α each time. Wait, but starting at P, the first chord goes from P to A1 to B1, which has a horizontal component of 4R cos² α as calculated before.Wait, in the first segment pair (P to A1 to B1), the horizontal progress is 4R cos² α. Then from B1 to A2 to B2, another horizontal progress. Let's compute that.From B1(R(4 cos² α -1), 0), going at angle α above PQ to A2, then back to B2. The horizontal progress from B1 to B2 would be similar to the progress from P to B1. Let's calculate the horizontal component.Using the same method as before, from point B1(x1, 0) = (R(4 cos² α -1), 0), the line at angle α above PQ is y = tan α (x - x1). Intersecting the circle x² + y² = R²:x² + tan² α (x - x1)^2 = R²Expand:x² + tan² α (x² - 2 x1 x + x1²) = R²(1 + tan² α)x² - 2 x1 tan² α x + tan² α x1² - R² = 0Multiply by cos² α:x² + sin² α (x² - 2 x1 x + x1²) - R² cos² α = 0Expand:x² + sin² α x² - 2 x1 sin² α x + sin² α x1² - R² cos² α = 0Combine like terms:x² (1 + sin² α) - 2 x1 sin² α x + sin² α x1² - R² cos² α = 0Again, since the line passes through B1(x1, 0), which is on the circle, x = x1 is a root. Let's factor it out:(x - x1)(something) = 0But let me compute the other root. Using quadratic formula:x = [2 x1 sin² α ± sqrt{4 x1² sin^4 α - 4 (1 + sin² α)(sin² α x1² - R² cos² α)}]/[2(1 + sin² α)]Simplify discriminant:4 x1² sin^4 α - 4 (1 + sin² α)(sin² α x1² - R² cos² α)= 4[ x1² sin^4 α - (1 + sin² α)(sin² α x1² - R² cos² α) ]= 4[ x1² sin^4 α - sin² α x1² (1 + sin² α) + R² cos² α (1 + sin² α) ]= 4[ x1² sin^4 α - x1² sin² α - x1² sin^4 α + R² cos² α (1 + sin² α) ]= 4[ - x1² sin² α + R² cos² α (1 + sin² α) ]So,x = [2 x1 sin² α ± sqrt{4(- x1² sin² α + R² cos² α (1 + sin² α))}]/[2(1 + sin² α)]Simplify:x = [x1 sin² α ± sqrt{ - x1² sin² α + R² cos² α (1 + sin² α) }]/(1 + sin² α)We know that one root is x1, so the other root (for A2) would be:x = [x1 sin² α - sqrt{ - x1² sin² α + R² cos² α (1 + sin² α) }]/(1 + sin² α)But this seems complicated. Let's substitute x1 = R(4 cos² α -1).Plugging in x1:sqrt{ - [R²(4 cos² α -1)^2 sin² α] + R² cos² α (1 + sin² α) }Factor out R²:R sqrt{ - (4 cos² α -1)^2 sin² α + cos² α (1 + sin² α) }This is getting too involved. Maybe there's a pattern here.Alternatively, notice that the horizontal progress after each pair of segments (up and down) forms a geometric sequence. The first progress is 4R cos² α, the next might be 4R cos² α * (2 cos 2α - 1), or something like that. However, without knowing the exact relation, it's hard to see.Alternatively, since after four peaks, the total horizontal displacement is 2R (from P to Q), maybe the sum of the horizontal components equals 2R. Each horizontal component alternates direction, but since we are moving from P to Q, all contributions are in the same direction.Wait, no. Actually, starting at P(-R, 0), the first horizontal progress is towards Q (positive direction). Each subsequent segment continues in the positive direction. Wait, no, because the path zig-zags, but overall it's progressing from P to Q. Wait, but in the first step, from P(-R, 0) to B1(R(4 cos² α -1), 0), the displacement is R(4 cos² α -1 +1) = 4R cos² α. Then from B1 to B2, the displacement would be 4R cos² α * (something). Wait, maybe each step multiplies the horizontal progress by 2 cos 2α. Let's see.If from P to B1 is 4R cos² α, then the next segment from B1 to B2 would be 4R cos² α * 2 cos 2α. Then total progress after two peaks would be 4R cos² α + 4R cos² α * 2 cos 2α. Wait, but this seems arbitrary.Alternatively, since each time you go up and down, the horizontal displacement is multiplied by 2 cos 2α. Let's recall that in the first step, the horizontal displacement is 4R cos² α. Then in the next step, from B1 to B2, we can think of it as a similar journey but starting from B1. The horizontal displacement would be 4R cos² α * 2 cos 2α. Then after four peaks, which is two such cycles, the total displacement would be 4R cos² α (1 + 2 cos 2α + (2 cos 2α)^2 + ... ). But this is speculative.Alternatively, think of this as a recurrence relation. Let’s denote d_n as the horizontal displacement after n peaks. We have d_1 = 4R cos² α. Then d_{n+1} = d_n + 4R cos² α * (2 cos 2α)^n. But I need to verify.Alternatively, recall that in similar problems, like light bouncing between two mirrors with an angle θ, the number of reflections needed to exit is related to θ. In this case, it's a circle and the path is constrained by angle α. After four reflections, it spans the diameter. Maybe using the concept of the least common multiple or something.Alternatively, think of the total rotation angle. Each time the path reflects off the circumference, it rotates by 2α. Wait, if the path makes angle α with PQ, then the angle between the incoming and outgoing path at the circumference is 2α. But how does this affect the overall rotation?Wait, at each peak (reflection), the direction of the path changes by 2α. If we start at P, after four reflections, the total rotation would be 4*2α = 8α. To end at Q, which is directly opposite P, the total rotation should be an odd multiple of 180°, but since we have four reflections, maybe it's 180°. So 8α = 180°, so α = 22.5°, but 22.5° is not one of the options. Hmm, conflicting.Alternatively, maybe the total change in direction is 180°, so 4*(π - 2α) = π. Wait, no, that might not be right. Wait, at each reflection, the direction changes by 2α. So after four reflections, the total change is 8α. To end up in the opposite direction (from P to Q), the total change should be π radians (180°). So 8α = π => α = π/8 = 22.5°, again not an option. So this suggests that my model is incorrect.Alternatively, perhaps the angle is measured differently. If the angle between the path and PQ is α, then the angle between the path and the tangent at the circumference is 90° - α. Then, by the law of reflection, the angle of incidence equals the angle of reflection. So the total change in direction at each peak is 2*(90° - α) = 180° - 2α. Therefore, after four peaks, the total rotation is 4*(180° - 2α). To end up going from P to Q, the total rotation should be 180°, so:4*(180° - 2α) = 180°=> 720° - 8α = 180°=> -8α = -540°=> α = 67.5°Still not in the options. Hmm.Wait, the options are 60°,72°,75°,80°,86°. 67.5 is not listed. So my previous approaches are not yielding an answer in the options. Maybe my model is incorrect.Alternative approach: Consider that the path consists of four equal chords, each making angle α with PQ. The total length of PQ is 2R. Each chord has a horizontal component. The sum of the horizontal components of the four chords equals 2R.But if there are four chords, each with horizontal component 2R cos α, then total horizontal would be 4*2R cos α = 8R cos α. But this equals 2R:8R cos α = 2R => cos α = 2/8 = 1/4 => α = arccos(1/4) ≈ 75.5°, which is close to option C (75°), but not exact. However, the options don't include 75.5°, but C is 75°. Maybe the approximation is expected. Alternatively, maybe there's a different factor.Wait, but perhaps the horizontal component per segment is different. From the coordinate calculation earlier, the horizontal progress after one peak (two segments) is 4R cos² α. So maybe after four peaks (eight segments), the total horizontal progress is 8R cos² α. But we need total progress to be 2R:8R cos² α = 2R => cos² α = 2/8 = 1/4 => cos α = 1/2 => α = 60°, which is option A.Ah! That seems promising. Let's verify.If each peak corresponds to two segments (up and down), then four peaks correspond to eight segments. However, the horizontal progress per peak (two segments) is 4R cos² α. Therefore, four peaks would give 4*(4R cos² α) = 16R cos² α. Wait, that can't be, since PQ is only 2R long.Wait, confusion here. Let's go back.Previously, we calculated that from P(-R, 0), after one peak (A1) and one crossing (B1), the horizontal progress was 4R cos² α. But PQ is only 2R in length. So 4R cos² α must be less than or equal to 2R. Therefore, cos² α ≤ 0.5, so α ≥ 45°. But our answer above, if we set 4R cos² α * 4 peaks = 2R, then 16R cos² α = 2R => cos² α = 1/8 => α ≈ 69°, which is still not in the options.Wait, perhaps each peak corresponds to one segment. Wait, the problem states "after four peaks, the zig-zag line ends at Q". So starting at P, the path goes up to peak1, down to crossing1, up to peak2, down to crossing2, up to peak3, down to crossing3, up to peak4, down to Q. So total segments: P-peak1, peak1-crossing1, crossing1-peak2, peak2-crossing2, crossing2-peak3, peak3-crossing3, crossing3-peak4, peak4-Q. That's eight segments, four peaks. Each peak is a high point (or low point) on the circumference.But in terms of horizontal progress, from P to crossing1 is the first horizontal step, then crossing1 to crossing2 is the second, etc. Wait, each "crossing" is a point where the path intersects PQ. So from P to crossing1 is the first horizontal step, crossing1 to crossing2 is the second, and so on. Since there are four peaks, there are four crossings between P and Q: crossing1, crossing2, crossing3, crossing4 (which is Q). Wait, but starting at P, then peaks and crossings:1. P to peak1 (on circumference)2. peak1 to crossing1 (on PQ)3. crossing1 to peak2 (on circumference)4. peak2 to crossing2 (on PQ)5. crossing2 to peak3 (on circumference)6. peak3 to crossing3 (on PQ)7. crossing3 to peak4 (on circumference)8. peak4 to Q (which is crossing4)So there are four crossings: crossing1, crossing2, crossing3, Q. Each between P and Q. So the horizontal progress is divided into four intervals: P to crossing1, crossing1 to crossing2, crossing2 to crossing3, crossing3 to Q.Each interval is covered by a pair of segments (up and down). For example, P to crossing1 is covered by P-peak1 and peak1-crossing1. The horizontal component of each interval is determined by the angle α.From earlier, the first interval (P to crossing1) has horizontal progress x1 = R(4 cos² α -1) - (-R) = R(4 cos² α -1 +1) = 4R cos² α. So each interval's horizontal progress is 4R cos² α. Then total horizontal progress after four intervals (four crossings) would be 4*(4R cos² α) = 16R cos² α. But since the total diameter is 2R, we set 16R cos² α = 2R => 16 cos² α = 2 => cos² α = 1/8 => cos α = √(1/8) = (√2)/4 ≈ 0.3535, so α ≈ 69.3°, which isn't an option. This contradicts.But earlier, when we calculated the first crossing point B1, we had x1 = R(4 cos² α -1). So starting from P at -R, the first crossing is at x1 = R(4 cos² α -1). The distance from P to B1 is x1 - (-R) = R(4 cos² α -1 +1) = 4R cos² α. Then from B1 to B2 would be similar. So each crossing point is determined by the previous crossing point. So the position after n crossings is x_n = -R + 4R cos² α + 4R cos² α*(2 cos 2α) + ... ?Wait, let's model the positions after each crossing. Let’s denote x0 = -R (position of P). After first crossing (B1), x1 = x0 + 4R cos² α. Then from B1 to B2, what's the horizontal progress?From B1 at x1, the path goes to peak2 at angle α, then to B2. The horizontal progress from B1 to B2 can be calculated similarly as 4R cos² α. But this would imply x2 = x1 + 4R cos² α. Then total progress after four crossings would be x4 = -R + 4*4R cos² α = -R + 16R cos² α. Setting x4 = R (position of Q):-R + 16R cos² α = R => 16R cos² α = 2R => 16 cos² α = 2 => cos² α = 1/8 => α ≈ 69.3°, still not matching the options.But the answer choices are 60°,72°,75°,80°,86°, none of which is approximately 69.3°. So either my approach is wrong, or I need to reconsider.Wait, perhaps the horizontal progress per crossing is not 4R cos² α, but something else. Let's recast the problem.If each segment from a crossing to a peak is a chord making angle α with PQ. The horizontal component of this chord is 2R cos α. But since you go up and then down, the total horizontal progress per peak is 2*(2R cos α)*cos α = 4R cos² α. Wait, no. Wait, each segment (up and down) is a chord. The horizontal component of each chord is 2R cos α. But going up and then down would cover 2*(2R cos α). But that would just be 4R cos α. But no, because the angle is maintained.Wait, maybe not. Let's think in terms of vectors. Each segment (from crossing to peak) is a vector with magnitude equal to the chord length and direction α relative to PQ. The chord length is 2R sin θ, where θ is the angle subtended at the center. But the angle α is the angle between the chord and PQ. So the relationship between θ and α is important.Alternatively, if a chord makes angle α with PQ (the diameter), then the angle between the chord and the radius at the point of incidence is 90° - α. By the law of reflection, the angle of incidence equals the angle of reflection. So the change in direction after each reflection is 2*(90° - α) = 180° - 2α. Thus, each peak changes the direction by 180° - 2α. After four peaks, the total direction change is 4*(180° - 2α). To end at Q, the total direction change should result in a straight line from P to Q, i.e., a change of 180°. So:4*(180° - 2α) = 180° + 360°*k, where k is an integer.The smallest positive solution is when k=0:720° - 8α = 180°=> 8α = 540°=> α = 67.5°, again not in the options. If k=1:720° -8α = 540°=> 8α = 180°=> α=22.5°, which is also not in the options. So this approach isn't yielding an answer in the given options.Alternative idea: Since after four peaks, the path connects P to Q, which is a diameter. The four peaks divide the circle into four equal arcs? If each peak is spaced equally around the circle, then the central angle between peaks would be 360°/4 = 90°, but the path is constrained by the angle α. Not sure.Alternatively, think of the points where the path meets the circumference. Each peak is a point on the circumference. The central angles between consecutive peaks would be related to α. If we can relate the central angle to α, then four central angles summing up to 180° (since PQ is a diameter). Wait, but the path starts at P, goes to A1, A2, A3, A4, and ends at Q. So the total central angle from P to A1 to A2 to A3 to A4 to Q should be 180°, as they are along the diameter.But each segment from a peak to the next is determined by the angle α. The chord from P to A1 makes angle α with PQ. Then the chord from A1 to A2 makes angle α with PQ, but on the other side. Wait, no, because after reflecting, the angle is maintained. This is getting too vague.Wait, going back to the coordinate system approach. From P(-R,0) to A1(R cos 2α, R sin 2α). Then from A1 to B1(R(4 cos² α -1),0). Then from B1 to A2: Let's calculate A2's coordinates.From B1(R(4 cos² α -1),0), the path makes angle α above PQ. The equation is y = tan α (x - R(4 cos² α -1)). To find A2, solve with the circle equation:x² + y² = R²Substitute y:x² + tan² α (x - R(4 cos² α -1))² = R²Expand:x² + tan² α (x² - 2 R(4 cos² α -1) x + R²(4 cos² α -1)^2 ) = R²As before, divide by cos² α:x² cos² α + sin² α (x² - 2 R(4 cos² α -1) x + R²(4 cos² α -1)^2 ) = R² cos² αExpand:x² cos² α + sin² α x² - 2 R sin² α (4 cos² α -1) x + sin² α R²(4 cos² α -1)^2 - R² cos² α = 0Combine like terms:x² (cos² α + sin² α) - 2 R sin² α (4 cos² α -1) x + R² [ sin² α (4 cos² α -1)^2 - cos² α ] = 0Simplify:x² - 2 R sin² α (4 cos² α -1) x + R² [ sin² α (16 cos^4 α -8 cos² α +1) - cos² α ] = 0This quadratic equation in x has roots at x = R(4 cos² α -1) (the point B1) and the new point A2's x-coordinate. Using Vieta's formula, the sum of roots is 2 R sin² α (4 cos² α -1). Therefore, the other root x2 = 2 R sin² α (4 cos² α -1) - R(4 cos² α -1)= R(4 cos² α -1)(2 sin² α -1)So x2 = R(4 cos² α -1)(2 sin² α -1)Similarly, the product of the roots is R² [ sin² α (16 cos^4 α -8 cos² α +1) - cos² α ].But since one root is x1 = R(4 cos² α -1), the other root x2 = [product]/x1:x2 = [ R² [ sin² α (16 cos^4 α -8 cos² α +1) - cos² α ] ] / [ R(4 cos² α -1) ]= R [ sin² α (16 cos^4 α -8 cos² α +1) - cos² α ] / (4 cos² α -1)This is complicated, but let's see if this can be simplified.Let’s compute the numerator:sin² α (16 cos^4 α -8 cos² α +1) - cos² α= 16 sin² α cos^4 α -8 sin² α cos² α + sin² α - cos² α= 16 sin² α cos^4 α -8 sin² α cos² α + (sin² α - cos² α)= 16 sin² α cos^4 α -8 sin² α cos² α - cos 2αThis seems too involved. Maybe there's a trigonometric identity that can simplify this.Alternatively, note that this approach is getting too algebraically intensive, and since the answer choices are all nice angles, perhaps we can substitute the options and see which one satisfies the condition that after four peaks, the path ends at Q.Let's test α = 60°.If α = 60°, then cos α = 0.5, sin α = √3/2.Compute the first crossing point B1:x1 = R(4 cos² α -1) = R(4*(0.5)^2 -1) = R(4*0.25 -1) = R(1 -1) = 0. So B1 is at the center O(0,0).Then from B1(0,0), the next segment at angle α=60° above PQ. The equation is y = tan 60° (x -0) = √3 x. Intersecting the circle x² + y² = R²:x² + 3x² = R² => 4x² = R² => x = ±R/2. Since we're moving towards Q, x = R/2. So A2 is at (R/2, (√3 R)/2). Then from A2, the path goes to B2 on PQ making angle α=60° below PQ. The equation is y - (√3 R)/2 = -√3 (x - R/2). Setting y=0:- (√3 R)/2 = -√3 (x - R/2)Divide both sides by -√3:R/2 = x - R/2 => x = R/2 + R/2 = R. So B2 is at Q(R, 0). But we were supposed to have four peaks, but here after two peaks (A1 and A2), we've already reached Q. So α=60° results in reaching Q after two peaks, which contradicts the problem's four peaks. Therefore, α=60° is incorrect.Next, test α=72°.Compute cos α = cos 72° ≈ 0.3090.First crossing point B1:x1 = R(4 cos² α -1). Compute cos² 72° ≈ (0.3090)^2 ≈ 0.0954.So x1 ≈ R(4*0.0954 -1) ≈ R(0.3816 -1) ≈ R(-0.6184). So B1 is at approximately -0.6184 R on PQ. But starting from P(-R,0), moving to A1 and then to B1≈-0.6184 R. This is moving towards Q but only a small distance. Then from B1, next segment at angle 72° above PQ. Compute A2.The line from B1≈-0.6184 R to A2 is y = tan72°(x +0.6184 R). tan72°≈3.0777.Intersect with circle:x² + y² = R²x² + [3.0777(x +0.6184 R)]² = R²This is messy to compute, but perhaps notice that the horizontal progress after four peaks would accumulate to 2R. However, manual calculation for each α is time-consuming. Alternatively, maybe there's a pattern.Alternatively, since the displacement after each peak follows a geometric progression with ratio k, such that the total displacement after four terms equals 2R. The first term is 4R cos² α, then each subsequent term is multiplied by (2 cos 2α -1). For example:Total displacement = 4R cos² α [1 + (2 cos 2α -1) + (2 cos 2α -1)^2 + (2 cos 2α -1)^3] = 2RBut this is speculative.Alternatively, if we assume that each reflection off the circumference causes the horizontal progress to be multiplied by a factor of 2 cos 2α, then after four peaks, the total progress is 4R cos² α * (1 + 2 cos 2α + (2 cos 2α)^2 + (2 cos 2α)^3). Setting this equal to 2R:4R cos² α * [1 + 2 cos 2α + 4 cos² 2α + 8 cos³ 2α] = 2RDivide both sides by R:4 cos² α * [1 + 2 cos 2α + 4 cos² 2α + 8 cos³ 2α] = 2This still seems complex, but let's test α=72°:Compute cos α ≈ 0.3090, cos² α ≈ 0.0954, cos 2α = 2 cos² α -1 ≈ 2*0.0954 -1 ≈ -0.8092.Then:4 *0.0954*[1 +2*(-0.8092)+4*(-0.8092)^2 +8*(-0.8092)^3]Compute inside the brackets:1 + 2*(-0.8092) ≈ 1 -1.6184 ≈ -0.61844*(-0.8092)^2 ≈4*0.6545≈2.6188*(-0.8092)^3≈8*(-0.5345)≈-4.276Total inside ≈ -0.6184 +2.618 -4.276≈-0.6184+2.618=2.0-4.276≈-2.276Multiply by 4*0.0954≈0.3816Total≈0.3816*(-2.276)≈-0.868, which is not equal to 2.Not matching. Test α=60°:cos α=0.5, cos 2α=2*(0.5)^2 -1= -0.5Inside the brackets:1 +2*(-0.5)+4*(-0.5)^2 +8*(-0.5)^3=1 -1 +4*0.25 +8*(-0.125)=0 +1 -1=0Thus total progress=0, which is incorrect.Test α=75°:cos α≈0.2588, cos²α≈0.06699, cos2α=2cos²α -1≈-0.8660Inside the brackets:1 +2*(-0.8660)+4*(-0.8660)^2 +8*(-0.8660)^3≈1 -1.732 +4*0.75 +8*(-0.6495)≈1 -1.732+3 -5.196≈(1 -1.732)+(3 -5.196)≈-0.732-2.196≈-2.928Multiply by 4*0.06699≈0.26796≈0.268Total≈0.268*(-2.928)≈-0.785, again not 2.Test α=80°:cos α≈0.1736, cos²α≈0.0301, cos2α=2*0.0301 -1≈-0.9398Inside the brackets:1 +2*(-0.9398)+4*(-0.9398)^2 +8*(-0.9398)^3≈1 -1.8796 +4*0.8836 +8*(-0.832)≈1 -1.8796+3.5344 -6.656≈(1 -1.8796)+(3.5344 -6.656)≈-0.8796 -3.1216≈-4.0012Multiply by 4*0.0301≈0.1204≈0.12Total≈0.12*(-4.0012)≈-0.48, nope.Test α=86°:cos α≈0.0698, cos²α≈0.00487, cos2α=2*0.00487 -1≈-0.99026Inside the brackets:1 +2*(-0.99026)+4*(-0.99026)^2 +8*(-0.99026)^3≈1 -1.9805 +4*0.9806 +8*(-0.9709)≈1 -1.9805 +3.9224 -7.7672≈(1 -1.9805)+(3.9224 -7.7672)≈-0.9805 -3.8448≈-4.8253Multiply by 4*0.00487≈0.0195Total≈0.0195*(-4.8253)≈-0.094, nope.None of the options seem to satisfy the equation. This suggests that my assumed relation is incorrect.Given that none of the approaches are working and the answer is among the options, perhaps I need to consider another method.Alternative approach: Consider that the problem is similar to a regular polygon inscribed in the circle. If there are four peaks, the path makes four equal chords subtending angles at the center. The angle α is related to the central angle.If after four peaks, the path goes from P to Q, which is a diameter (180°). If each peak corresponds to a equal arc length, then the central angle between consecutive peaks is 180°/4 = 45°, but this is a guess.If each peak subtends an angle of θ at the center, then four peaks would sum to 180°, so θ = 45°. The angle α is the angle between the chord and the diameter. For a chord subtending θ at the center, the angle between the chord and the diameter is α = (180° - θ)/2 = (180° -45°)/2 = 67.5°, which is not an option. Hence, this approach is invalid.Alternatively, if the four peaks divide the semicircle into five equal parts (since four peaks create five segments), but that would be 180°/5 = 36°, leading to α = (180° -36°)/2=72°, which is option B. This could be the answer.Wait, let me explain. If the semicircle from P to Q is 180°, and the four peaks divide it into five equal arcs, each arc is 36°. Then each chord between two consecutive peaks subtends 36° at the center. The angle α between the chord and the diameter PQ would be half of the arc's supplementary angle. Wait, the angle between a chord and the diameter is equal to half the measure of the arc it subtends. Wait, actually, the angle between a chord and the diameter is equal to half the measure of the arc opposite to that angle.Wait, in a circle, the angle between a chord and the diameter at the point where they meet is equal to half the measure of the arc opposite to the angle. So if the chord subtends an arc of 72°, then the angle between the chord and the diameter is 36°. But in our case, if each peak divides the semicircle into five parts, each arc is 36°, so the angle α would be 36°, which is not in the options. Hmm, conflicting.Alternatively, if the four peaks divide the semicircle into four equal arcs, each 45°, then the angle α would be 45°, which is not in the options.Alternatively, if the angle α corresponds to the angle of a regular pentagon (72°), which is option B. Since 72° is a common angle in regular polygons, and the problem involves four peaks, which might relate to five divisions (since four peaks create five segments). So if the semicircle is divided into five equal parts, each 36°, then the angle α would be 72°.Wait, this seems like a stretch, but if the angle α is formed by a chord that subtends an arc of 144°, then the angle between the chord and the diameter is 72°, which is option B. That could make sense. Let's verify.The angle between a chord and the diameter: If a chord subtends an arc of 2θ at the center, then the angle between the chord and the diameter is θ. For example, if the chord subtends 120°, the angle between the chord and the diameter is 60°. So if α =72°, then the arc subtended by the chord is 2*72°=144°. Four such arcs would cover 4*144°=576°, which is more than the semicircle's 180°, so that can't be.Alternatively, the total change in direction after each reflection is 180° - 2α, and after four reflections, it should amount to 180°, so:4*(180° - 2α) = 180°Which gives α=67.5°, not in options. But if the total change should be 360°, then:4*(180° -2α)=360°180° -2α=90°α=45°, which is not an option.Alternatively, total change should be 540°, which is 1.5 full rotations:4*(180° -2α)=540°180°-2α=135°2α=45°α=22.5°, nope.Alternatively, considering the problem might involve external angles in a polygon. If the path forms a star polygon with four points, but that’s unclear.Given that all my calculations point towards α≈67.5°, which is not an option, but the closest option is 60° and 72°, but 72° is in the options. Maybe the problem expects an approximate answer or there's a different interpretation.Wait, going back to the coordinate system approach with α=72°:First crossing point B1:x1 = R(4 cos² 72° -1). cos72≈0.3090, cos²72≈0.0954, 4*0.0954=0.3816, 0.3816-1=-0.6184. So B1 is at x=-0.6184 R.Then from B1, the path goes at angle 72° above PQ. Equation: y=tan72°*(x +0.6184 R). tan72≈3.0777.Intersect with circle:x² + y²=R²x² + [3.0777(x +0.6184 R)]² = R²Approximate computation:Let’s compute at R=1 for simplicity.x² + [3.0777x +1.903]^2 =1Expand:x² +9.464x² +11.703x +3.621=110.464x² +11.703x +2.621=0Solving:x = [-11.703 ±sqrt(11.703² -4*10.464*2.621)]/(2*10.464)Discriminant≈136.96 -109.9≈27.06sqrt(27.06)≈5.202x≈[-11.703 ±5.202]/20.928x1≈(-11.703+5.202)/20.928≈-6.501/20.928≈-0.31x2≈(-11.703-5.202)/20.928≈-16.905/20.928≈-0.807But B1 is at x≈-0.6184. So the other intersection point is x≈-0.31. So A2 is at x≈-0.31, y≈3.0777*(-0.31 +0.6184)≈3.0777*0.3084≈0.949. So A2≈(-0.31,0.949). Then from A2, the path goes to B2 at angle 72° below PQ.Equation: y -0.949 = -tan72°(x +0.31)Set y=0:-0.949 = -3.0777(x +0.31)0.949/3.0777≈0.308≈x +0.31x≈0.308 -0.31≈-0.002So B2 is at x≈-0.002, approximately 0. Then from B2, the path goes to A3 at angle72° above PQ. The line is y=tan72°*(x +0.002). Intersecting the circle:x² + [3.0777(x +0.002)]²=1x² +9.464x² +0.0123x +0.000012=110.464x² +0.0123x -0.999988=0Solutions≈x≈[ -0.0123 ±sqrt(0.000151 +41.81) ]/20.928≈sqrt(41.81)=6.466≈(6.466)/20.928≈0.309. So x≈0.309, y≈3.0777*(0.309+0.002)≈3.0777*0.311≈0.956. So A3≈(0.309,0.956). Then from A3, path goes to B3 at angle72° below PQ:Equation: y -0.956 = -3.0777(x -0.309). Set y=0:-0.956 = -3.0777x +0.951-1.907 = -3.0777x => x≈0.619. Then from B3(0.619,0), path goes to A4 at angle72° above PQ: y=3.0777(x -0.619). Intersect circle:x² + [3.0777(x -0.619)]²=1x² +9.464(x² -1.238x +0.619²)=1x² +9.464x² -11.703x +3.621=110.464x² -11.703x +2.621=0Which is the same equation as before, leading to x≈0.618 or x≈0.31. But x≈0.618 is Q at x=1? Wait, no. For R=1, Q is at x=1. But solving 10.464x² -11.703x +2.621=0:Discriminant≈136.96 -109.9≈27.06x=(11.703±5.202)/20.928x1≈(11.703+5.202)/20.928≈16.905/20.928≈0.807x2≈(11.703-5.202)/20.928≈6.501/20.928≈0.31So x≈0.807 or 0.31. Since moving from B3(0.619,0) towards Q, the next intersection is at x≈0.807, y≈3.0777*(0.807 -0.619)=3.0777*0.188≈0.579. So A4≈(0.807,0.579). Then from A4, path goes to B4 making angle72° below PQ:Equation: y -0.579 = -3.0777(x -0.807). Set y=0:-0.579 = -3.0777x +2.483-3.0777x = -0.579 -2.483 = -3.062x≈3.062/3.0777≈0.995. So B4≈(0.995,0), which is almost Q(1,0). After four peaks (A1, A2, A3, A4), the path reaches approximately Q. So α=72° gets us close to Q after four peaks. Thus, α=72° is likely the correct answer, option B.This trial-and-error with α=72° showed that after four peaks, the path reaches very close to Q. Given the options, 72° is the most plausible answer.Final Answerboxed{B}</think>Given a circle with diameter (PQ), a zig-zag line starts at point (P) and ends at point (Q) after four peaks. Each angle between the zig-zag line and the diameter (PQ) is (alpha). We need to determine the value of (alpha).1. Understanding the Problem: - The zig-zag line starts at (P) and ends at (Q) after four peaks. - Each angle between the zig-zag line and the diameter (PQ) is (alpha). - The key is to determine the angle (alpha) such that the zig-zag line, after four peaks, spans the diameter (PQ).2. Coordinate System Approach: - Place the circle in a coordinate system with (P) at ((-R, 0)) and (Q) at ((R, 0)). - The first segment from (P) makes an angle (alpha) with (PQ) and intersects the circumference at a point (A_1). - Using trigonometry and solving the quadratic equation for the circle, the coordinates of (A_1) are found to be ((R cos 2alpha, R sin 2alpha)).3. Subsequent Segments: - The path continues from (A_1) to intersect (PQ) at a point (B_1), calculated as (R(4 cos^2 alpha - 1)). - This process repeats for subsequent peaks and intersections. The horizontal progress after each segment is analyzed.4. Testing Answer Choices: - Testing (alpha = 72^circ) (option B): - The first crossing point (B_1) is calculated. - Subsequent segments are computed, showing that after four peaks, the path reaches very close to (Q).5. Conclusion: - After detailed calculations and verification, it is found that (alpha = 72^circ) results in the zig-zag line ending at (Q) after four peaks.Thus, the size of angle (alpha) is (boxed{B}).]

🔑:SolUtionBAfter four peaks the zig-zag is at the end of the diameter, so after two peaks it must be at the centre O of the circle. The triangle O P R is isosceles since O P and O R are both radii, hence angle P R O=alpha and angle P O R=180^{circ}-2 alpha. Angle O T R=180^{circ}-alpha (angles on a straight line), and hence angle T R O=3 alpha-180^{circ} (angles in triangle T R O ).Triangle O T S is isosceles because its base angles are equal, and hence S T=S O. Therefore triangles O P R and OTS are congruent because they both have two sides which equal the radius of the circle, with angle 180^{circ}-2 alpha between (SAS). Hence O T=P R=T R (since P R=R T in isosceles triangle P R T ). Hence triangle O T R is isosceles, and its base angles are equal, that is 180^{circ}-2 alpha=3 alpha-180^{circ}. Solving this gives alpha=72^{circ}.

🧠:[<think>Okay, so I need to find the maximum and minimum values of the function y = (2x² + 6x + 6)/(x² + 4x + 5). Hmm, let's see. I remember that to find extrema of a function, you take its derivative and set it equal to zero, then solve for x. Then check those points to see if they're maxima or minima. But since this is a rational function, maybe there's another way too? Like maybe simplifying it or completing the square? Let me think.First, let me check if the denominator can be factored or simplified. The denominator is x² + 4x + 5. Let's try to complete the square. x² + 4x + 4 + 1 = (x + 2)² + 1. So the denominator is always positive because (x + 2)² is non-negative and adding 1 makes it at least 1. So the function is defined for all real numbers, no vertical asymptotes.Similarly, maybe the numerator can be completed as well. The numerator is 2x² + 6x + 6. Let me factor out a 2 first: 2(x² + 3x) + 6. Completing the square for x² + 3x: take half of 3, which is 1.5, square it to get 2.25. So x² + 3x + 2.25 - 2.25 = (x + 1.5)² - 2.25. Then multiply by 2: 2[(x + 1.5)² - 2.25] + 6 = 2(x + 1.5)² - 4.5 + 6 = 2(x + 1.5)² + 1.5. So the numerator is 2(x + 1.5)² + 1.5, which is also always positive since it's a square term plus 1.5. So both numerator and denominator are always positive, meaning the function y is always positive. But that might not help directly with maxima or minima.Alternatively, maybe I can perform polynomial division or express this as a constant plus a fraction. Let's try that. Let me write y = (2x² + 6x + 6)/(x² + 4x + 5). Let's divide numerator and denominator by x² + 4x + 5. So, how many times does x² + 4x + 5 go into 2x² + 6x + 6? Well, 2 times. Because 2*(x² + 4x +5) = 2x² +8x +10. Then subtract that from the numerator: (2x² +6x +6) - (2x² +8x +10) = -2x -4. So y = 2 + (-2x -4)/(x² +4x +5). So y = 2 - (2x +4)/(x² +4x +5). Hmm, that simplifies it a bit. Maybe that helps with finding the maximum and minimum? Let's see.Alternatively, maybe set y = (2x² +6x +6)/(x² +4x +5) and solve for x in terms of y. Let's try cross-multiplying: y(x² +4x +5) = 2x² +6x +6. Then expand the left side: yx² +4yx +5y = 2x² +6x +6. Bring all terms to one side: yx² +4yx +5y -2x² -6x -6 =0. Combine like terms: (y -2)x² + (4y -6)x + (5y -6) =0. This is a quadratic equation in x. For real solutions to exist, the discriminant must be non-negative. So discriminant D = [4y -6]^2 -4*(y -2)*(5y -6) ≥0.Let me compute D: (4y -6)^2 -4*(y -2)*(5y -6). First expand (4y -6)^2: 16y² -48y +36. Then compute 4*(y -2)*(5y -6): 4*(5y² -6y -10y +12) =4*(5y² -16y +12)=20y² -64y +48. Now subtract the two: (16y² -48y +36) - (20y² -64y +48) =16y² -48y +36 -20y² +64y -48= (-4y² +16y -12). So discriminant D = -4y² +16y -12 ≥0.Multiply both sides by -1 (which reverses the inequality): 4y² -16y +12 ≤0. Divide both sides by 4: y² -4y +3 ≤0. Factor the quadratic: y² -4y +3 = (y -1)(y -3) ≤0. So the inequality holds when y is between 1 and 3, inclusive. Therefore, the range of y is [1,3], so the minimum value is 1 and maximum is 3. Therefore, the maximum is 3 and minimum is 1. Wait, is that correct? Let me verify.But maybe I made a mistake in the discriminant. Let me check again. So discriminant D = (4y -6)^2 -4*(y -2)*(5y -6). Let me compute each part step by step. (4y -6)^2 = 16y² -48y +36. 4*(y -2)*(5y -6): first compute (y -2)*(5y -6) =5y² -6y -10y +12=5y² -16y +12. Then multiply by 4: 20y² -64y +48. So D =16y² -48y +36 -20y² +64y -48= (16y² -20y²) + (-48y +64y) + (36 -48)= -4y² +16y -12. So that's correct. Then D ≥0 implies -4y² +16y -12 ≥0. Divide both sides by -4: y² -4y +3 ≤0. Which factors as (y -1)(y -3) ≤0. Therefore, y is between 1 and 3. So that's correct.Therefore, the minimum value is 1 and maximum is 3. So the function's range is [1,3]. Therefore, the answer is maximum 3, minimum 1. But wait, maybe I should check if these values are actually attained. Let's see. When does y =1? Let me set y=1 and solve for x.From the equation y=(2x² +6x +6)/(x² +4x +5). If y=1, then 1=(2x² +6x +6)/(x² +4x +5). Multiply both sides by denominator: x² +4x +5 = 2x² +6x +6. Bring all terms to left: x² +4x +5 -2x² -6x -6= -x² -2x -1=0. Multiply by -1: x² +2x +1=0. Which is (x +1)^2=0, so x= -1. So at x=-1, y=1. Similarly, when y=3, set 3=(2x² +6x +6)/(x² +4x +5). Multiply both sides: 3x² +12x +15=2x² +6x +6. Bring all terms to left: 3x² +12x +15 -2x² -6x -6= x² +6x +9=0. Which is (x +3)^2=0, so x= -3. Therefore, at x= -3, y=3. Therefore, both y=1 and y=3 are attained at x=-1 and x=-3, respectively. So those are the minimum and maximum points.Alternatively, if I use calculus, take derivative of y and set to zero. Let me try that method to confirm.First, y=(2x² +6x +6)/(x² +4x +5). Let’s compute derivative using quotient rule: y’= [ (denominator)(numerator)’ - numerator*(denominator)’ ] / (denominator)^2.Compute numerator derivative: numerator is 2x² +6x +6, so derivative is 4x +6.Denominator derivative: denominator is x² +4x +5, so derivative is 2x +4.Therefore, y’= [ (x² +4x +5)(4x +6) - (2x² +6x +6)(2x +4) ] / (x² +4x +5)^2.Let me compute the numerator:First term: (x² +4x +5)(4x +6). Let me expand that:x²*(4x +6) +4x*(4x +6) +5*(4x +6)=4x³ +6x² +16x² +24x +20x +30=4x³ + (6x² +16x²)=22x² + (24x +20x)=44x +30.Second term: (2x² +6x +6)(2x +4). Let's expand that:2x²*(2x +4) +6x*(2x +4) +6*(2x +4)=4x³ +8x² +12x² +24x +12x +24=4x³ + (8x² +12x²)=20x² + (24x +12x)=36x +24.So numerator is first term minus second term: (4x³ +22x² +44x +30) - (4x³ +20x² +36x +24)= (4x³ -4x³)+(22x² -20x²)+(44x -36x)+(30 -24)= 0 +2x² +8x +6=2x² +8x +6.Therefore, y’= (2x² +8x +6)/(x² +4x +5)^2.Set numerator equal to zero: 2x² +8x +6=0. Divide by 2: x² +4x +3=0. Factor: (x +1)(x +3)=0. So x=-1 and x=-3. So critical points at x=-1 and x=-3. Then evaluate y at these points.At x=-1: y=(2*(-1)^2 +6*(-1)+6)/( (-1)^2 +4*(-1)+5 )= (2*1 -6 +6)/(1 -4 +5)= (2 -6 +6)/(2)=2/2=1.At x=-3: y=(2*9 +6*(-3)+6)/(9 +4*(-3)+5)= (18 -18 +6)/(9 -12 +5)=6/2=3.So confirms previous result. So maximum at x=-3, y=3; minimum at x=-1, y=1.Therefore, the maximum is 3 and minimum is 1.Wait, but maybe check the second derivative to confirm concavity? Just to be thorough. Although since there are only two critical points and the function approaches a horizontal asymptote, which is y=2 (since degrees are equal, leading coefficients 2/1=2), so as x approaches infinity, y approaches 2. So since the function approaches 2 at both ends, and we found a maximum at 3 and minimum at 1, which are above and below 2, that makes sense.Alternatively, using the earlier method where we rewrote y as 2 - (2x +4)/(x² +4x +5). Let me see if that helps in analyzing the maximum and minimum. So y = 2 - (2x +4)/(x² +4x +5). Let me denote the fraction as A = (2x +4)/(x² +4x +5). Then y = 2 - A. So to find maximum and minimum of y, we need to find the minimum and maximum of A, respectively. Because if A is maximized, then y is minimized (since it's 2 - A), and if A is minimized, y is maximized.So let's analyze A = (2x +4)/(x² +4x +5). Let's compute derivative of A:A’ = [ (x² +4x +5)(2) - (2x +4)(2x +4) ] / (x² +4x +5)^2.Compute numerator: 2(x² +4x +5) - (2x +4)^2. Let's compute each term:First term: 2x² +8x +10.Second term: (2x +4)^2=4x² +16x +16.So numerator: 2x² +8x +10 -4x² -16x -16= (-2x² -8x -6)= -2(x² +4x +3)= -2(x +1)(x +3).Set numerator equal to zero: -2(x +1)(x +3)=0. So critical points at x=-1 and x=-3. So A has critical points at x=-1 and x=-3.Evaluate A at x=-1: (2*(-1) +4)/( (-1)^2 +4*(-1)+5 )= (-2 +4)/(1 -4 +5)=2/2=1. So A=1 here, so y=2 -1=1.At x=-3: (2*(-3) +4)/(9 +4*(-3)+5)= (-6 +4)/(9 -12 +5)= (-2)/2= -1. So A=-1 here, so y=2 - (-1)=3.Therefore, when A is maximized at 1, y is minimized at 1; when A is minimized at -1, y is maximized at 3. Therefore, confirms the previous results.So whether using the discriminant method, calculus with quotient rule, or rewriting the function and analyzing the fraction, all methods point to the maximum being 3 and minimum being 1.Therefore, the answer is maximum 3 and minimum 1.Final AnswerThe maximum of the function is boxed{3} and the minimum is boxed{1}.</think>To determine the maximum and minimum values of the function y = frac{2x^2 + 6x + 6}{x^2 + 4x + 5},we can use the method of setting the discriminant of a quadratic equation to be non-negative. First, cross-multiplying to express the equation in terms of ( y ):y(x^2 + 4x + 5) = 2x^2 + 6x + 6.Rearranging terms, we get a quadratic equation in ( x ):(y - 2)x^2 + (4y - 6)x + (5y - 6) = 0.For this quadratic equation to have real solutions, the discriminant must be non-negative:(4y - 6)^2 - 4(y - 2)(5y - 6) geq 0.Simplifying the discriminant:16y^2 - 48y + 36 - 4(5y^2 - 16y + 12) geq 0,which simplifies to:-4y^2 + 16y - 12 geq 0.Dividing by -4 and factoring:(y - 1)(y - 3) leq 0.This inequality holds when ( 1 leq y leq 3 ). Therefore, the range of the function is [1, 3]. To verify, we check the critical points using calculus. The derivative of ( y ) is found using the quotient rule:y' = frac{(x^2 + 4x + 5)(4x + 6) - (2x^2 + 6x + 6)(2x + 4)}{(x^2 + 4x + 5)^2}.Simplifying the numerator:2x^2 + 8x + 6 = 0,which factors to:(x + 1)(x + 3) = 0.The critical points are ( x = -1 ) and ( x = -3 ). Evaluating ( y ) at these points:- At ( x = -1 ), ( y = 1 ).- At ( x = -3 ), ( y = 3 ).Thus, the function attains its minimum value at ( x = -1 ) and its maximum value at ( x = -3 ). Final AnswerThe maximum of the function is boxed{3} and the minimum is boxed{1}.,<think>Okay, so I need to find the maximum and minimum values of the function y = (2x² + 6x + 6)/(x² + 4x + 5). Hmm, let's see. I remember that for functions like this, which are rational functions where both the numerator and denominator are quadratics, one way to find maxima and minima is to take the derivative and set it equal to zero. But maybe there's another method too, like completing the square or something? Hmm. Let me think.First, maybe I can simplify the function. Let me see if the denominator can be factored or completed into a square. The denominator is x² + 4x + 5. To complete the square, take half of 4, which is 2, square it to get 4, so x² + 4x + 4 + 1 = (x + 2)² + 1. So that's always positive, right? Since (x + 2)² is non-negative and adding 1 makes it at least 1. So the denominator is never zero, which is good. That means the function is defined for all real numbers. So no vertical asymptotes.What about horizontal asymptotes? Since both numerator and denominator are degree 2, the horizontal asymptote will be the ratio of the leading coefficients, which is 2/1 = 2. So as x approaches ±∞, y approaches 2. That might be useful. So maybe the maximum or minimum is near that horizontal asymptote?But to find exact maxima and minima, I think calculus is the way to go. Let's recall that to find extrema, we take the derivative of y with respect to x, set it equal to zero, and solve for x. Then check those points to see if they're maxima or minima.So let's compute y'. Since y is a quotient of two functions, we can use the quotient rule. The quotient rule says that if y = f/g, then y' = (f'g - fg') / g².Let me write that down:f(x) = 2x² + 6x + 6g(x) = x² + 4x + 5Then f'(x) = 4x + 6g'(x) = 2x + 4So applying the quotient rule:y' = [ (4x + 6)(x² + 4x + 5) - (2x² + 6x + 6)(2x + 4) ] / (x² + 4x + 5)^2Okay, now I need to expand the numerator and simplify it. Let's do that step by step.First, compute (4x + 6)(x² + 4x + 5):Multiply 4x by each term: 4x³ + 16x² + 20xMultiply 6 by each term: 6x² + 24x + 30So adding those together: 4x³ + 16x² + 20x + 6x² + 24x + 30Combine like terms:4x³ + (16x² + 6x²) = 4x³ + 22x²(20x + 24x) = 44xAnd then +30So total is 4x³ + 22x² + 44x + 30Next, compute (2x² + 6x + 6)(2x + 4):Multiply 2x² by each term: 4x³ + 8x²Multiply 6x by each term: 12x² + 24xMultiply 6 by each term: 12x + 24Add them all up: 4x³ + 8x² + 12x² + 24x + 12x + 24Combine like terms:4x³ + (8x² + 12x²) = 4x³ + 20x²(24x + 12x) = 36x+24So total is 4x³ + 20x² + 36x + 24Now, subtract the second result from the first:Numerator = (4x³ + 22x² + 44x + 30) - (4x³ + 20x² + 36x + 24)Let's subtract term by term:4x³ - 4x³ = 022x² - 20x² = 2x²44x - 36x = 8x30 - 24 = 6So the numerator simplifies to 2x² + 8x + 6.Therefore, the derivative y' is:y' = (2x² + 8x + 6) / (x² + 4x + 5)^2Now, to find critical points, set y' = 0:(2x² + 8x + 6) / (x² + 4x + 5)^2 = 0Since the denominator is always positive (as we saw earlier, it's (x + 2)^2 + 1), the fraction is zero when the numerator is zero. So set:2x² + 8x + 6 = 0Divide both sides by 2:x² + 4x + 3 = 0Factor:(x + 1)(x + 3) = 0Thus, x = -1 or x = -3So these are the critical points. Now, we need to determine whether these points correspond to maxima or minima. To do that, we can use the second derivative test or analyze the sign changes of the first derivative around these points. Alternatively, since there are only two critical points, we can just plug them into the original function and compare the values.Let me compute y at x = -1 and x = -3.First, x = -1:y = [2(-1)^2 + 6(-1) + 6] / [(-1)^2 + 4(-1) + 5] = [2(1) -6 +6] / [1 -4 +5] = [2 -6 +6]/[2] = 2/2 = 1Then x = -3:y = [2(-3)^2 + 6(-3) + 6] / [(-3)^2 + 4(-3) + 5] = [2(9) -18 +6]/[9 -12 +5] = [18 -18 +6]/[2] = 6/2 = 3So at x = -1, y = 1; at x = -3, y = 3.Since we have a horizontal asymptote at y = 2, which is between 1 and 3. So when x approaches infinity, y approaches 2 from below because the leading coefficient ratio is 2/1 = 2, but the numerator has a slightly smaller coefficient? Wait, both numerator and denominator are 2x² and x². Wait, leading terms are 2x² / x² = 2. So as x approaches ±infinity, y approaches 2. But depending on the lower terms, maybe it approaches from above or below.Wait, let's check when x is very large positive, say x = 1000:Numerator: 2*(1000)^2 + 6*1000 +6 ≈ 2,000,000 + 6,000 +6 ≈ 2,006,006Denominator: (1000)^2 +4*1000 +5 ≈ 1,000,000 +4,000 +5 ≈ 1,004,005So y ≈ 2,006,006 / 1,004,005 ≈ approximately 2.0. Let's compute it more precisely:2,006,006 / 1,004,005 = (2*1,004,005 + 2,006,006 - 2*1,004,005) / 1,004,005Wait, that might not help. Alternatively, approximate:Divide numerator and denominator by 1000^2:Numerator ≈ 2 + 6/1000 + 6/(1000)^2 ≈ 2.006006Denominator ≈ 1 + 4/1000 +5/(1000)^2 ≈ 1.004005So y ≈ 2.006006 / 1.004005 ≈ (2.006006 / 1.004005) ≈ approximately 2.006 / 1.004 ≈ Let's divide 2.006 by 1.004.Multiply numerator and denominator by 1000: 2006 / 1004 ≈ 1.998. Hmm, so approximately 1.998, which is just below 2. So as x approaches positive infinity, y approaches 2 from below.Similarly, if x is a large negative number, say x = -1000:Numerator: 2*(-1000)^2 +6*(-1000) +6 = 2,000,000 -6,000 +6 ≈ 1,994,006Denominator: (-1000)^2 +4*(-1000) +5 = 1,000,000 -4,000 +5 ≈ 996,005So y ≈ 1,994,006 / 996,005 ≈ approximately 2.0. Let's approximate:Divide numerator and denominator by 1000^2:Numerator ≈ 2 -6/1000 +6/(1000)^2 ≈ 1.994006Denominator ≈ 1 -4/1000 +5/(1000)^2 ≈ 0.996005So y ≈ 1.994006 / 0.996005 ≈ approximately 2.002. So that's slightly above 2. So as x approaches negative infinity, y approaches 2 from above.So, putting this together, the function approaches 2 from above as x approaches -infty and from below as x approaches +infty. The critical points we found are at x = -3 (y = 3) and x = -1 (y = 1). So between these two points, the function must go from 3 down to 1, but considering the horizontal asymptote, maybe there's a maximum at x = -3 and a minimum at x = -1.Wait, but let's check the behavior around these points. Let's test the derivative sign.We have critical points at x = -3 and x = -1. Let's choose test points in the intervals:1. x < -3, say x = -42. -3 < x < -1, say x = -23. x > -1, say x = 0Compute the sign of y' (numerator) in each interval.First, the numerator of y' is 2x² +8x +6. Let's factor that: 2(x² +4x +3) = 2(x +1)(x +3). So the quadratic crosses zero at x = -1 and x = -3. Since the coefficient of x² is positive, the parabola opens upwards.Therefore, the sign of the numerator (and hence y') is:- For x < -3: Let's pick x = -4: 2*(-4)^2 +8*(-4) +6 = 32 -32 +6 = 6 >0- For -3 < x < -1: x = -2: 2*4 + (-16) +6 = 8 -16 +6 = -2 <0- For x > -1: x = 0: 0 +0 +6 =6 >0Therefore, the derivative is positive when x < -3, negative between -3 and -1, and positive again when x > -1.Therefore, the function is increasing when x < -3, decreasing between -3 and -1, and increasing again when x > -1.Therefore, x = -3 is a local maximum (since the function goes from increasing to decreasing there), and x = -1 is a local minimum (since the function goes from decreasing to increasing there).Therefore, the maximum value of y is 3 at x = -3, and the minimum value is 1 at x = -1.Wait, but we need to confirm that these are indeed the global maxima and minima. Since the function approaches 2 at both infinities, and the maximum at x = -3 is 3 which is higher than 2, and the minimum at x = -1 is 1 which is lower than 2, so yes, those are the global maximum and minimum.Alternatively, since there are only two critical points and the function tends to 2 at both ends, which is between 1 and 3, so the highest point is 3 and the lowest is 1.Therefore, the answer is maximum 3 and minimum 1.Wait, let me double-check by plugging in another point. For example, at x = -2, let's compute y:y = [2*(-2)^2 +6*(-2)+6]/[(-2)^2 +4*(-2)+5] = [8 -12 +6]/[4 -8 +5] = [2]/[1] = 2. So at x = -2, y = 2. That makes sense. So between x = -3 and x = -1, the function decreases from 3 to 1, passing through y = 2 at x = -2. Then as x increases past -1, it starts increasing again towards the horizontal asymptote y = 2 from below.Similarly, as x approaches -infty, the function approaches 2 from above, reaches a peak at x = -3 (y=3), then decreases to the minimum at x = -1 (y=1), and then increases back towards 2 as x approaches +infty. So yeah, that seems consistent.So I think that's it. The maximum is 3 and the minimum is 1.Final AnswerThe maximum of the function is boxed{3} and the minimum is boxed{1}.</think>To determine the maximum and minimum values of the function ( y = frac{2x^2 + 6x + 6}{x^2 + 4x + 5} ), we start by finding the critical points using calculus.First, we compute the derivative using the quotient rule. Let ( f(x) = 2x^2 + 6x + 6 ) and ( g(x) = x^2 + 4x + 5 ). The derivatives are ( f'(x) = 4x + 6 ) and ( g'(x) = 2x + 4 ). Applying the quotient rule:[y' = frac{(4x + 6)(x^2 + 4x + 5) - (2x^2 + 6x + 6)(2x + 4)}{(x^2 + 4x + 5)^2}]Expanding and simplifying the numerator:[(4x + 6)(x^2 + 4x + 5) = 4x^3 + 22x^2 + 44x + 30][(2x^2 + 6x + 6)(2x + 4) = 4x^3 + 20x^2 + 36x + 24][text{Numerator} = (4x^3 + 22x^2 + 44x + 30) - (4x^3 + 20x^2 + 36x + 24) = 2x^2 + 8x + 6]Thus, the derivative is:[y' = frac{2x^2 + 8x + 6}{(x^2 + 4x + 5)^2}]Setting the numerator equal to zero to find critical points:[2x^2 + 8x + 6 = 0 implies x^2 + 4x + 3 = 0 implies (x + 1)(x + 3) = 0 implies x = -1 text{ or } x = -3]Evaluating the function at these critical points:- At ( x = -1 ): [ y = frac{2(-1)^2 + 6(-1) + 6}{(-1)^2 + 4(-1) + 5} = frac{2 - 6 + 6}{1 - 4 + 5} = frac{2}{2} = 1 ]- At ( x = -3 ): [ y = frac{2(-3)^2 + 6(-3) + 6}{(-3)^2 + 4(-3) + 5} = frac{18 - 18 + 6}{9 - 12 + 5} = frac{6}{2} = 3 ]Analyzing the sign of the derivative around the critical points, we find that the function is increasing for ( x < -3 ), decreasing between ( -3 ) and ( -1 ), and increasing again for ( x > -1 ). This indicates a local maximum at ( x = -3 ) and a local minimum at ( x = -1 ).Given the horizontal asymptote ( y = 2 ), the maximum value ( 3 ) and minimum value ( 1 ) are indeed global.Thus, the maximum of the function is (boxed{3}) and the minimum is (boxed{1}).]

🔑:I. solution: Express x in terms of y from the given relationship. This way, we can specify which y values allow for the calculation of x, in other words, which y numbers have an x value for which our function takes the value y, i.e., what is the range of our function. From this, we can select the desired maximum or minimum - if there is a largest or smallest number in the range.The denominator is never 0 because x^{2}+4 x+5=(x+2)^{2}+1 geq 1, so the relationship obtained by cross-multiplication and rearrangement(y-2) x^{2}+(4 y-6) x+(5 y-6)=0is equivalent to the original. From this,x=frac{-4 y+6 pm sqrt{-4 y^{2}+16 y-12}}{2(y-2)}provided y neq 2, and x=-2 if y=2.x is given by the quadratic formula only if the discriminant is non-negative, i.e., if - immediately factoring the quadratic expression --4(y-1)(y-3) geq 0This is clearly true if and only if1 leq y leq 3This double inequality describes the entire range of our function, as the y=2 value obtained by a different route also satisfies it. According to this, our function has both a maximum and a minimum, namelyy_{max }=3, quad text { or } quad y_{min }=1values. These are taken by the function atx_{max }=-3, quad text { or } quad x_{min }=-1points.Remark. Some solutions arrived at the correct result with the following erroneous reasoning: "The maximum and minimum values are those which the function takes only once, these are where the discriminant is 0." According to this, every value of a linear function would be both a maximum and a minimum; furthermore, the value y=1 would not be a maximum of the y=sin x function, nor would y=-1 be its minimum, because each value is taken infinitely many times. This shows that the concept of extreme values is based solely on quadratic functions.II. solution. By division and completing the square in the denominator, the given expression can be transformed as follows:y=2-frac{2 x+4}{x^{2}+4 x+5}=2-frac{2(x+2)}{(x+2)^{2}+1}and by substituting x+2 with a new variable zy=2-frac{2 z}{z^{2}+1}=2-2 y_{1}, quad text { where } quad y_{1}=frac{z}{z^{2}+1}According to this, y has a maximum where the function y_{1} takes its minimum value, and a minimum where y_{1} is maximal. y_{1} can be positive, negative, and 0, because it has the same sign as the numerator z, since its denominator is positive, and y_{1}=0 when z=0; therefore, the maximum value of y_{1} can only be positive, and its minimum value can only be negative. It is sufficient to examine y_{1} for positive z values, because the values it takes for z and -z differ only in sign, so the negative of the maximum value to be determined, and the location of the minimum value, will give the minimum value and location.Further transformation givesy_{1}=frac{z}{z^{2}+1}=frac{1}{z+frac{1}{z}}=frac{1}{y_{2}}, quad text { where } quad y_{2}=z+frac{1}{z}so the maximum of y_{1} is equal to the minimum of the function y_{2}. Note that for positive z, the geometric mean of the two terms of y_{2} is 1, a constant. It is also known that the arithmetic mean of positive numbers is never less than their geometric mean, and is equal to it when the numbers are equal. Therefore, the minimum of y_{2} is 1, so y_{2, min }=2, and this occurs when z is the positive number for which z=1 / z, i.e., z=1.Now the maximum of y_{1} is 1 / 2, and according to the above, the minimum is y_{1, min }=-1 / 2 at z=-1; finally, according to (2), the minimum of y is 1, the maximum is 3, and these extreme values are taken at x=z-2 considering the above, at x=-1, and x=-3.III. solution: y can also be transformed as follows:begin{aligned}& y=1+frac{x^{2}+2 x+1}{x^{2}+4 x+5}=1+frac{(x+1)^{2}}{(x+2)^{2}+1} & y=3-frac{x^{2}+6 x+9}{x_{2}+4 x+5}=3-frac{(x+3)^{2}}{(x+2)^{2}+1}end{aligned}The second term of the first form is never negative, because its numerator is a perfect square and its denominator is positive. Therefore, y takes its minimum value at x+1=0, i.e., at x=-1; here the second term is 0, and the value of y is 1. Similarly, the subtrahend of the second form is smallest when x+3=0, i.e., x=-3, and here its value is 0, so the maximum of y is 3.